{"id":119074,"date":"2022-05-02T16:39:14","date_gmt":"2022-05-02T11:09:14","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=119074"},"modified":"2022-05-02T16:39:52","modified_gmt":"2022-05-02T11:09:52","slug":"chapter-4-quadratic-equations-questions-and-answers-ncert-solutions-for-class-10-mathematics","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/education\/chapter-4-quadratic-equations-questions-and-answers-ncert-solutions-for-class-10-mathematics","title":{"rendered":"Chapter 4 – Quadratic Equations Questions and Answers: NCERT Solutions for Class 10 Mathematics"},"content":{"rendered":"

Exercise 4.1<\/h2>\n

1. Check whether the following are quadratic equations:
\n(i) (x + 1)2\u00a0= 2(x \u2013 3)
\n(ii) x2\u00a0\u2013 2x = (\u20132) (3 \u2013 x)
\n(iii) (x \u2013 2)(x + 1) = (x \u2013 1)(x + 3)
\n(iv) (x \u2013 3)(2x +1) = x(x + 5)
\n(v) (2x \u2013 1)(x \u2013 3) = (x + 5)(x \u2013 1)
\n(vi) x2\u00a0+ 3x + 1 = (x \u2013 2)2
\n(vii) (x + 2)3\u00a0= 2x (x2\u00a0\u2013 1)
\n(viii) x3\u00a0\u2013 4×2\u00a0\u2013 x + 1 = (x \u2013 2)3<\/h2>\n

Solutions:
\n(i) Given,
\n(x + 1)2\u00a0= 2(x \u2013 3)
\nBy using the formula for (a+b)2\u00a0= a2+2ab+b2
\n\u21d2\u00a0x2\u00a0+ 2x\u00a0+ 1 = 2x\u00a0\u2013 6
\n\u21d2\u00a0x2\u00a0+ 7 = 0
\nSince the above equation is in the form of ax2\u00a0+\u00a0bx\u00a0+\u00a0c\u00a0= 0.
\nTherefore, the given equation is\u00a0quadratic equation.<\/h3>\n

(ii) Given, x2\u00a0\u2013 2x = (\u20132) (3 \u2013 x)
\n\u21d2\u00a0x2\u00a0\u2013\u00a02x = -6\u00a0+ 2x
\n\u21d2\u00a0x2\u00a0\u2013 4x\u00a0+ 6 = 0
\nSince the above equation is in the form of ax2\u00a0+\u00a0bx\u00a0+\u00a0c\u00a0= 0.
\nTherefore, the given equation is\u00a0quadratic equation.<\/h3>\n

(iii) Given, (x \u2013 2)(x + 1) = (x \u2013 1)(x + 3)
\nBy multiplication
\n\u21d2\u00a0x2\u00a0\u2013\u00a0x\u00a0\u2013 2 =\u00a0x2\u00a0+ 2x\u00a0\u2013 3
\n\u21d2 3x\u00a0\u2013 1 = 0
\nSince the above equation is not in the form of ax2\u00a0+\u00a0bx\u00a0+\u00a0c\u00a0= 0.
\nTherefore, the given equation is\u00a0not a quadratic equation.<\/h3>\n

(iv) Given, (x \u2013 3)(2x +1) = x(x + 5)
\nBy multiplication
\n\u21d2 2×2\u00a0\u2013 5x\u00a0\u2013 3 =\u00a0x2\u00a0+ 5x
\n\u21d2 \u00a0x2\u00a0\u2013 10x\u00a0\u2013 3 = 0
\nSince the above equation is in the form of ax2\u00a0+\u00a0bx\u00a0+\u00a0c\u00a0= 0.
\nTherefore, the given equation is\u00a0quadratic equation.<\/h3>\n

(v) Given, (2x\u00a0\u2013 1)(x\u00a0\u2013 3) = (x\u00a0+ 5)(x\u00a0\u2013 1)
\nBy multiplication
\n\u21d2 2×2\u00a0\u2013 7x\u00a0+\u00a03 =\u00a0x2\u00a0+ 4x\u00a0\u2013 5
\n\u21d2\u00a0x2\u00a0\u2013 11x\u00a0+\u00a08 = 0
\nSince the above equation is in the form of ax2\u00a0+\u00a0bx\u00a0+\u00a0c\u00a0= 0.
\nTherefore, the given equation is\u00a0quadratic equation.<\/h3>\n

(vi) Given, x2\u00a0+ 3x\u00a0+ 1 = (x\u00a0\u2013 2)2
\nBy using the formula for (a-b)2=a2-2ab+b2
\n\u21d2\u00a0x2\u00a0+ 3x\u00a0+ 1\u00a0=\u00a0x2\u00a0+ 4\u00a0\u2013 4x
\n\u21d2 7x\u00a0\u2013 3 = 0
\nSince the above equation is not in the form of ax2\u00a0+\u00a0bx\u00a0+\u00a0c\u00a0= 0.
\nTherefore, the given equation is\u00a0not a quadratic equation.<\/h3>\n

(vii) Given, (x\u00a0+ 2)3\u00a0= 2x(x2\u00a0\u2013 1)
\nBy using the formula for (a+b)3\u00a0= a3+b3+3ab(a+b)
\n\u21d2\u00a0x3\u00a0+ 8\u00a0+\u00a0x2\u00a0+ 12x\u00a0= 2×3\u00a0\u2013 2x
\n\u21d2\u00a0x3\u00a0+ 14x\u00a0\u2013 6×2\u00a0\u2013 8 = 0
\nSince the above equation is not in the form of ax2\u00a0+\u00a0bx\u00a0+\u00a0c\u00a0= 0.
\nTherefore, the given equation is\u00a0not a quadratic equation.<\/h3>\n

(viii) Given,\u00a0x3\u00a0\u2013 4×2\u00a0\u2013\u00a0x\u00a0+ 1 = (x\u00a0\u2013 2)3
\nBy using the formula for (a-b)3\u00a0= a3-b3-3ab(a-b)
\n\u21d2 \u00a0x3\u00a0\u2013 4×2\u00a0\u2013\u00a0x\u00a0+ 1\u00a0=\u00a0x3\u00a0\u2013 8 \u2013 6×2\u00a0\u00a0+ 12x
\n\u21d2 2×2\u00a0\u2013 13x\u00a0+ 9 = 0
\nSince the above equation is in the form of ax2\u00a0+\u00a0bx\u00a0+\u00a0c\u00a0= 0.
\nTherefore, the given equation is\u00a0quadratic equation.<\/h3>\n

2. Represent the following situations in the form of quadratic equations:
\n(i) The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
\n(ii) The product of two consecutive positive integers is 306. We need to find the integers.
\n(iii) Rohan\u2019s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan\u2019s present age.
\n(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km\/h less, then it would have taken<\/h2>\n

Solutions:
\n(i) Let us consider,
\nBreadth of the rectangular plot =\u00a0x\u00a0m
\nThus, the length of the plot = (2x\u00a0+ 1) m.
\nAs we know,
\nArea of rectangle = length\u00a0\u00d7\u00a0breadth = 528 m2
\nPutting the value of length and breadth of the plot in the formula, we get,
\n(2x\u00a0+ 1) \u00d7\u00a0x\u00a0= 528
\n\u21d2 2×2\u00a0+\u00a0x\u00a0=528
\n\u21d2 2×2\u00a0+\u00a0x\u00a0\u2013 528 = 0
\nTherefore, the length and breadth of plot, satisfies the quadratic equation, 2×2\u00a0+\u00a0x\u00a0\u2013 528 = 0, which is the required representation of the problem mathematically.<\/h3>\n

(ii) Let us consider,
\nThe first integer number =\u00a0x
\nThus, the next consecutive positive integer will be =\u00a0x\u00a0+ 1
\nProduct of two consecutive integers =\u00a0x\u00a0\u00d7\u00a0(x\u00a0+1) = 306
\n\u21d2\u00a0x2\u00a0+\u00a0x\u00a0= 306
\n\u21d2\u00a0x2\u00a0+\u00a0x\u00a0\u2013 306 = 0
\nTherefore, the two integers x and x+1, satisfies the quadratic equation, x2\u00a0+\u00a0x\u00a0\u2013 306 = 0, which is the required representation of the problem mathematically.<\/h3>\n

(iii) Let us consider,
\nAge of Rohan\u2019s =\u00a0x\u00a0 years
\nTherefore, as per the given question,
\nRohan\u2019s mother\u2019s age =\u00a0x\u00a0+ 26
\nAfter 3 years,
\nAge of Rohan\u2019s =\u00a0x\u00a0+ 3
\nAge of Rohan\u2019s mother will be =\u00a0x\u00a0+ 26 + 3 =\u00a0x\u00a0+ 29
\nThe product of their ages after 3 years will be equal to 360, such that
\n(x\u00a0+ 3)(x\u00a0+ 29) = 360
\n\u21d2\u00a0x2\u00a0+ 29x\u00a0+ 3x\u00a0+ 87 = 360
\n\u21d2\u00a0x2\u00a0+ 32x\u00a0+ 87 \u2013 360 = 0
\n\u21d2\u00a0x2\u00a0+ 32x\u00a0\u2013 273 = 0
\nTherefore, the age of Rohan and his mother, satisfies the quadratic equation, x2\u00a0+ 32x\u00a0\u2013 273 = 0, which is the required representation of the problem mathematically.<\/h3>\n

(iv) Let us consider,
\nThe speed of train =\u00a0x\u00a0 km\/h
\nAnd
\nTime taken to travel 480 km = 480\/x km\/hr
\nAs per second condition, the speed of train = (x\u00a0\u2013 8) km\/h
\nAlso given, the train will take 3 hours to cover the same distance.
\nTherefore, time taken to travel 480 km = (480\/x)+3 km\/h
\nAs we know,
\nSpeed \u00d7 Time = Distance
\nTherefore,
\n(x\u00a0\u2013 8)(480\/x\u00a0)+ 3 = 480
\n\u21d2 480\u00a0+ 3x\u00a0\u2013 3840\/x\u00a0\u2013 24 = 480
\n\u21d2 3x\u00a0\u2013 3840\/x\u00a0= 24
\n\u21d2\u00a0x2\u00a0\u2013 8x\u00a0\u2013 1280 = 0
\nTherefore, the speed of the train, satisfies the quadratic equation,\u00a0x2\u00a0\u2013 8x\u00a0\u2013 1280 = 0, which is the required representation of the problem mathematically.<\/h3>\n

Exercise 4.2<\/h2>\n

1. Find the roots of the following quadratic equations by factorisation:
\n(i)\u00a0x2\u00a0\u2013 3x\u00a0\u2013 10 = 0
\n(ii) 2×2\u00a0+\u00a0x\u00a0\u2013 6 = 0
\n(iii) \u221a2\u00a0x2\u00a0+ 7x\u00a0+ 5\u221a2\u00a0= 0
\n(iv) 2×2\u00a0\u2013\u00a0x\u00a0+1\/8 = 0
\n(v) 100×2\u00a0\u2013 20x\u00a0+ 1 = 0<\/h2>\n

Solutions:
\n(i) Given,\u00a0x2\u00a0\u2013 3x\u00a0\u2013 10 =0
\nTaking LHS,
\n=>x2\u00a0\u2013 5x\u00a0+ 2x\u00a0\u2013 10
\n=>x(x\u00a0\u2013 5)\u00a0+ 2(x\u00a0\u2013 5)
\n=>(x\u00a0\u2013 5)(x\u00a0+\u00a02)
\nThe roots of this equation,\u00a0x2\u00a0\u2013 3x\u00a0\u2013 10 = 0 are the values of x for which\u00a0(x\u00a0\u2013 5)(x\u00a0+\u00a02) = 0
\nTherefore,\u00a0x\u00a0\u2013 5 = 0 or\u00a0x\u00a0+ 2 = 0
\n=>\u00a0x\u00a0= 5 or\u00a0x\u00a0= -2<\/h3>\n

(ii) Given, 2×2\u00a0+\u00a0x\u00a0\u2013 6 = 0
\nTaking LHS,
\n=> 2×2\u00a0+ 4x\u00a0\u2013 3x\u00a0\u2013 6
\n=> 2x(x\u00a0+\u00a02) \u2013 3(x\u00a0+ 2)
\n=> (x\u00a0+ 2)(2x\u00a0\u2013 3)
\nThe roots of this equation, 2×2\u00a0+\u00a0x\u00a0\u2013 6=0 are the values of x for which\u00a0(x\u00a0x\u00a0+ 2)(2x\u00a0\u2013 3)\u00a0= 0
\nTherefore,\u00a0x\u00a0+ 2\u00a0= 0 or\u00a02x\u00a0\u2013 3 = 0
\n=>\u00a0x\u00a0= -2 or\u00a0x\u00a0= 3\/2<\/h3>\n

(iii) \u221a2\u00a0x2\u00a0+ 7x\u00a0+ 5\u221a2=0
\nTaking LHS,
\n=> \u221a2\u00a0x2\u00a0+ 5x\u00a0+ 2x\u00a0+ 5\u221a2
\n=>\u00a0x\u00a0(\u221a2x\u00a0+ 5)\u00a0+\u00a0\u221a2(\u221a2x\u00a0+ 5)= (\u221a2x\u00a0+ 5)(x\u00a0+\u00a0\u221a2)
\nThe roots of this equation, \u221a2\u00a0x2\u00a0+ 7x\u00a0+ 5\u221a2=0 are the values of x for which\u00a0(\u221a2x\u00a0+ 5)(x\u00a0+\u00a0\u221a2)\u00a0= 0
\nTherefore, \u221a2x\u00a0+ 5\u00a0= 0 or\u00a0x\u00a0+\u00a0\u221a2\u00a0= 0
\n=>\u00a0x\u00a0= -5\/\u221a2\u00a0or\u00a0x\u00a0= -\u221a2<\/h3>\n

(iv) 2×2\u00a0\u2013\u00a0x\u00a0+1\/8 = 0
\nTaking LHS,
\n=1\/8 (16×2 \u00a0\u2013 8x\u00a0+ 1)
\n= 1\/8 (16×2 \u00a0\u2013 4x\u00a0-4x\u00a0+ 1)
\n= 1\/8 (4x(4x\u00a0\u00a0\u2013 1) -1(4x\u00a0\u2013\u00a01))
\n= 1\/8 (4x\u00a0\u2013 1)2
\nThe roots of this equation, 2×2\u00a0\u2013\u00a0x\u00a0+ 1\/8 = 0, are the values of x for which\u00a0(4x\u00a0\u2013 1)2= 0
\nTherefore, (4x\u00a0\u2013 1) = 0 or (4x\u00a0\u2013 1) = 0
\n\u21d2\u00a0x\u00a0= 1\/4 or\u00a0x\u00a0= 1\/4<\/h3>\n

(v) Given, 100×2\u00a0\u2013 20x\u00a0+ 1=0
\nTaking LHS,
\n= 100×2\u00a0\u2013 10x\u00a0\u2013 10x\u00a0+ 1
\n= 10x(10x\u00a0\u2013 1) -1(10x\u00a0\u2013 1)
\n= (10x\u00a0\u2013 1)2
\nThe roots of this equation, 100×2\u00a0\u2013 20x\u00a0+ 1=0, are the values of x for which\u00a0(10x\u00a0\u2013 1)2= 0
\n\u2234 (10x\u00a0\u2013 1) = 0 or (10x\u00a0\u2013 1) = 0
\n\u21d2x = 1\/10 or x = 1\/10<\/h3>\n

2. Solve the problems given in Example 1.
\nRepresent the following situations mathematically:
\n(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
\n(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ` 750. We would like to find out the number of toys produced on that day.<\/h2>\n

Solutions:
\n(i) Let us say, the number of marbles John have =\u00a0x.
\nTherefore, number of marble Jivanti have = 45 \u2013\u00a0x
\nAfter losing 5 marbles each,
\nNumber of marbles John have =\u00a0x\u00a0\u2013 5
\nNumber of marble Jivanti have = 45 \u2013\u00a0x\u00a0\u2013 5 = 40 \u2013\u00a0x
\nGiven that the product of their marbles is 124.
\n\u2234 (x\u00a0\u2013 5)(40 \u2013\u00a0x) = 124
\n\u21d2\u00a0x2\u00a0\u2013 45x\u00a0+ 324 = 0
\n\u21d2\u00a0x2\u00a0\u2013 36x\u00a0\u2013 9x\u00a0+ 324 = 0
\n\u21d2\u00a0x(x\u00a0\u2013 36) -9(x\u00a0\u2013 36) = 0
\n\u21d2 (x\u00a0\u2013 36)(x\u00a0\u2013 9) = 0
\nThus, we can say,
\nx\u00a0\u2013 36 = 0 or\u00a0x\u00a0\u2013 9 = 0
\n\u21d2\u00a0x\u00a0= 36 or\u00a0x\u00a0= 9
\nTherefore,
\nIf, John\u2019s marbles = 36,
\nThen, Jivanti\u2019s marbles = 45 \u2013 36 = 9
\nAnd if John\u2019s marbles = 9,
\nThen, Jivanti\u2019s marbles = 45 \u2013 9 = 36<\/h3>\n

(ii) Let us say, number of toys produced in a day be\u00a0x.
\nTherefore, cost of production of each toy = Rs(55 \u2013\u00a0x)
\nGiven, total cost of production of the toys = Rs 750
\n\u2234\u00a0x(55 \u2013\u00a0x) = 750
\n\u21d2\u00a0x2\u00a0\u2013\u00a055x\u00a0+ 750 = 0
\n\u21d2\u00a0x2\u00a0\u2013 25x\u00a0\u2013\u00a030x\u00a0+ 750 = 0
\n\u21d2\u00a0x(x\u00a0\u2013 25) -30(x\u00a0\u2013 25) = 0
\n\u21d2 (x\u00a0\u2013 25)(x\u00a0\u2013 30) = 0
\nThus, either\u00a0x\u00a0-25 = 0 or\u00a0x\u00a0\u2013 30 = 0
\n\u21d2\u00a0x\u00a0= 25 or\u00a0x\u00a0= 30
\nHence, the number of toys produced in a day, will be either 25 or 30.<\/h3>\n

3. Find two numbers whose sum is 27 and product is 182.<\/h2>\n

Solution:
\nLet us say, first number be\u00a0x\u00a0and the second number is 27 \u2013\u00a0x.
\nTherefore, the product of two numbers
\nx(27 \u2013\u00a0x) = 182
\n\u21d2\u00a0x2\u00a0\u2013 27x\u00a0\u2013 182 = 0
\n\u21d2\u00a0x2\u00a0\u2013 13x\u00a0\u2013 14x\u00a0+ 182 = 0
\n\u21d2\u00a0x(x\u00a0\u2013 13) -14(x\u00a0\u2013 13) = 0
\n\u21d2 (x\u00a0\u2013 13)(x\u00a0-14) = 0
\nThus, either, x\u00a0= -13\u00a0= 0 or\u00a0x\u00a0\u2013 14 = 0
\n\u21d2\u00a0x\u00a0= 13 or\u00a0x\u00a0= 14
\nTherefore, if first number = 13, then second number = 27 \u2013 13 = 14
\nAnd if first number = 14, then second number = 27 \u2013 14 = 13
\nHence, the numbers are 13 and 14.<\/h3>\n

4. Find two consecutive positive integers, sum of whose squares is 365.<\/h2>\n

Solution:
\nLet us say, the two consecutive positive integers be\u00a0x\u00a0and\u00a0x\u00a0+ 1.
\nTherefore, as per the given questions,
\nx2\u00a0+ (x\u00a0+ 1)2\u00a0= 365
\n\u21d2\u00a0x2\u00a0+\u00a0x2\u00a0+ 1 + 2x\u00a0= 365
\n\u21d2 2×2\u00a0+ 2x \u2013 364 = 0
\n\u21d2\u00a0x2\u00a0+\u00a0x\u00a0\u2013 182 = 0
\n\u21d2\u00a0x2\u00a0+ 14x\u00a0\u2013 13x\u00a0\u2013 182 = 0
\n\u21d2\u00a0x(x\u00a0+\u00a014) -13(x\u00a0+\u00a014) = 0
\n\u21d2 (x\u00a0+ 14)(x\u00a0\u2013 13) = 0
\nThus, either,\u00a0x\u00a0+ 14 = 0 or\u00a0x\u00a0\u2013 13 = 0,
\n\u21d2\u00a0x\u00a0= \u2013 14 or\u00a0x\u00a0= 13
\nsince, the integers are positive,\u00a0so\u00a0x\u00a0can be 13, only.
\n\u2234\u00a0x\u00a0+ 1 = 13 + 1 = 14
\nTherefore, two consecutive positive integers will be 13 and 14.<\/h3>\n

5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.<\/h2>\n

Solution:
\nLet us say, the base of the right triangle be\u00a0x\u00a0cm.
\nGiven, the altitude of right triangle = (x\u00a0\u2013 7) cm
\nFrom Pythagoras theorem, we know,
\nBase2\u00a0+ Altitude2\u00a0= Hypotenuse2
\n\u2234\u00a0x2\u00a0+ (x\u00a0\u2013 7)2\u00a0= 132
\n\u21d2\u00a0x2\u00a0+\u00a0x2\u00a0+ 49 \u2013 14x\u00a0= 169
\n\u21d2 2×2\u00a0\u2013 14x\u00a0\u2013 120 = 0
\n\u21d2\u00a0x2\u00a0\u2013 7x\u00a0\u2013 60 = 0
\n\u21d2\u00a0x2\u00a0\u2013 12x\u00a0+ 5x\u00a0\u2013 60 = 0
\n\u21d2\u00a0x(x\u00a0\u2013 12) + 5(x\u00a0\u2013 12) = 0
\n\u21d2 (x\u00a0\u2013 12)(x\u00a0+ 5) = 0
\nThus, either\u00a0x\u00a0\u2013 12 = 0 or\u00a0x\u00a0+ 5 = 0,
\n\u21d2\u00a0x\u00a0= 12 or\u00a0x\u00a0= \u2013 5
\nSince sides cannot be negative,\u00a0x\u00a0can only be 12.
\nTherefore, the base of the given triangle is 12 cm and the altitude of this triangle will be (12 \u2013 7) cm = 5 cm.<\/h3>\n

6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs.90, find the number of articles produced and the cost of each article.<\/h2>\n

Solution:
\nLet us say, the number of articles produced be\u00a0x.
\nTherefore, cost of production of each article = Rs (2x\u00a0+ 3)
\nGiven, total cost of production is Rs.90
\n\u2234\u00a0x(2x\u00a0+ 3) = 90
\n\u21d2 2×2\u00a0+ 3x\u00a0\u2013 90 = 0
\n\u21d2 2×2\u00a0+ 15x\u00a0-12x\u00a0\u2013 90 = 0
\n\u21d2\u00a0x(2x\u00a0+ 15) -6(2x\u00a0+ 15) = 0
\n\u21d2 (2x\u00a0+ 15)(x\u00a0\u2013 6) = 0
\nThus, either 2x\u00a0+ 15 = 0 or\u00a0x\u00a0\u2013 6 = 0
\n\u21d2\u00a0x\u00a0= -15\/2 or\u00a0x\u00a0= 6
\nAs the number of articles produced can only be a positive integer, therefore, x can only be 6.
\nHence, number of articles produced = 6
\nCost of each article = 2 \u00d7 6 + 3 = Rs 15.<\/h3>\n

Exercise 4.3<\/h2>\n

1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
\n(i) 2×2\u00a0\u2013 7x\u00a0+3 = 0
\n(ii)\u00a02×2\u00a0+\u00a0x\u00a0\u2013 4 = 0
\n(iii)\u00a04×2\u00a0+ 4\u221a3x\u00a0+ 3 = 0
\n(iv)\u00a02×2\u00a0+\u00a0x\u00a0+ 4 = 0<\/h2>\n

Solutions:
\n(i) 2×2\u00a0\u2013\u00a07x\u00a0+ 3 = 0
\n\u21d2 2×2\u00a0\u2013\u00a07x\u00a0= \u2013 3
\nDividing by 2 on both sides, we get
\n\u21d2 x2\u00a0-7x\/2 = -3\/2
\n\u21d2 x2\u00a0-2 \u00d7 x \u00d77\/4 = -3\/2
\nOn adding (7\/4)2\u00a0to both sides of equation, we get
\n\u21d2 (x)2-2\u00d7x\u00d77\/4 +(7\/4)2\u00a0= (7\/4)2-3\/2
\n\u21d2 (x-7\/4)2\u00a0= (49\/16) \u2013 (3\/2)
\n\u21d2(x-7\/4)2\u00a0= 25\/16
\n\u21d2(x-7\/4)2\u00a0= \u00b15\/4
\n\u21d2\u00a0x\u00a0= 7\/4 \u00b1 5\/4
\n\u21d2\u00a0x\u00a0= 7\/4 + 5\/4 or x = 7\/4 \u2013 5\/4
\n\u21d2 x = 12\/4 or x = 2\/4
\n\u21d2\u00a0x = 3 or x = 1\/2<\/h3>\n

(ii) 2×2\u00a0+\u00a0x\u00a0\u2013 4 = 0
\n\u21d2 2×2\u00a0+\u00a0x\u00a0= 4
\nDividing both sides of the equation by 2, we get
\n\u21d2\u00a0x2\u00a0+x\/2 = 2
\nNow on adding (1\/4)2\u00a0to both sides of the equation, we get,
\n\u21d2 (x)2\u00a0+\u00a02 \u00d7\u00a0x\u00a0\u00d7 1\/4 + (1\/4)2\u00a0= 2\u00a0+ (1\/4)2
\n\u21d2 (x\u00a0+ 1\/4)2\u00a0= 33\/16
\n\u21d2\u00a0x\u00a0+ 1\/4 = \u00b1 \u221a33\/4
\n\u21d2\u00a0x\u00a0= \u00b1 \u221a33\/4 \u2013 1\/4
\n\u21d2\u00a0x\u00a0= (\u00b1 \u221a33-1)\/4
\nTherefore, either x\u00a0= (\u221a33-1)\/4 or\u00a0x\u00a0= (-\u221a33-1)\/4<\/h3>\n

(iii) 4×2\u00a0+ 4\u221a3x\u00a0+ 3 = 0
\nConverting the equation into a2+2ab+b2\u00a0form, we get,
\n\u21d2 (2x)2\u00a0+ 2 \u00d7 2x\u00a0\u00d7 \u221a3\u00a0+ (\u221a3)2\u00a0= 0
\n\u21d2 (2x\u00a0+ \u221a3)2\u00a0= 0
\n\u21d2 (2x\u00a0+ \u221a3) = 0 and (2x\u00a0+ \u221a3) = 0
\nTherefore, either\u00a0x\u00a0= -\u221a3\/2 or\u00a0x\u00a0= -\u221a3\/2.<\/h3>\n

(iv) 2×2\u00a0+\u00a0x\u00a0+ 4 = 0
\n\u21d2 2×2\u00a0+\u00a0x\u00a0= -4
\nDividing both sides of the equation by 2, we get
\n\u21d2\u00a0x2\u00a0+ 1\/2x\u00a0= 2
\n\u21d2\u00a0x2\u00a0+ 2\u00a0\u00d7\u00a0x\u00a0\u00d7 1\/4 = -2
\nBy adding (1\/4)2\u00a0to both sides of the equation, we get
\n\u21d2 (x)2\u00a0+\u00a02 \u00d7\u00a0x\u00a0\u00d7 1\/4 + (1\/4)2\u00a0= (1\/4)2\u00a0\u2013 2
\n\u21d2 (x\u00a0+ 1\/4)2\u00a0= 1\/16 \u2013 2
\n\u21d2 (x\u00a0+ 1\/4)2\u00a0= -31\/16
\nAs we know, the square of numbers cannot be negative.
\nTherefore, there is no real root for the given equation, 2×2\u00a0+\u00a0x\u00a0+ 4 = 0.<\/h3>\n

2. Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.
\n(i) 2×2\u00a0\u2013\u00a07x\u00a0+ 3 = 0
\nOn comparing the given equation with\u00a0ax2\u00a0+\u00a0bx\u00a0+\u00a0c\u00a0= 0, we get,
\na\u00a0= 2,\u00a0b\u00a0= -7 and\u00a0c\u00a0= 3
\nBy using quadratic formula, we get,<\/h2>\n

\"\"
\n\u21d2\u00a0x\u00a0= (7\u00b1\u221a(49 \u2013 24))\/4
\n\u21d2\u00a0x\u00a0= (7\u00b1\u221a25)\/4
\n\u21d2\u00a0x\u00a0= (7\u00b15)\/4
\n\u21d2\u00a0x\u00a0= (7+5)\/4 or\u00a0x\u00a0= (7-5)\/4
\n\u21d2\u00a0x\u00a0= 12\/4 or 2\/4
\n\u2234 \u00a0x\u00a0=\u00a03 or 1\/2<\/h3>\n

(ii) 2×2\u00a0+\u00a0x\u00a0\u2013 4 = 0<\/h2>\n

On comparing the given equation with\u00a0ax2\u00a0+\u00a0bx\u00a0+\u00a0c\u00a0= 0, we get,
\na = 2, b = 1 and\u00a0c\u00a0= -4
\nBy using quadratic formula, we get,<\/h3>\n

\"\"
\n\u21d2x\u00a0= (-1\u00b1\u221a1+32)\/4
\n\u21d2x\u00a0= (-1\u00b1\u221a33)\/4
\n\u2234\u00a0x\u00a0= (-1+\u221a33)\/4 or\u00a0x\u00a0= (-1-\u221a33)\/4<\/h3>\n

(iii) 4×2\u00a0+\u00a04\u221a3x\u00a0+ 3 = 0<\/h2>\n

On comparing the given equation with\u00a0ax2\u00a0+\u00a0bx\u00a0+\u00a0c\u00a0= 0, we get
\na\u00a0=\u00a04,\u00a0b\u00a0=\u00a04\u221a3\u00a0and\u00a0c = 3
\nBy using quadratic formula, we get,<\/h3>\n

\"\"
\n\u21d2\u00a0x\u00a0= (-4\u221a3\u00b1\u221a48-48)\/8
\n\u21d2\u00a0x\u00a0= (-4\u221a3\u00b10)\/8
\n\u2234\u00a0x\u00a0= -\u221a3\/2 or\u00a0x\u00a0= -\u221a3\/2<\/h3>\n

(iv)\u00a02×2\u00a0+\u00a0x\u00a0+ 4 = 0<\/h2>\n

On comparing the given equation with\u00a0ax2\u00a0+\u00a0bx\u00a0+\u00a0c\u00a0= 0, we get,
\na\u00a0= 2,\u00a0b\u00a0= 1 and\u00a0c\u00a0= 4
\nBy using quadratic formula, we get<\/h3>\n

\"\"
\n\u21d2\u00a0x\u00a0= (-1\u00b1\u221a1-32)\/4
\n\u21d2\u00a0x\u00a0= (-1\u00b1\u221a-31)\/4
\nAs we know, the square of a number can never be negative. Therefore, there is no real solution for the given equation.<\/h3>\n

3. Find the roots of the following equations:
\n(i)\u00a0x-1\/x\u00a0= 3,\u00a0x\u00a0\u2260 0
\n(ii) 1\/x+4 \u2013 1\/x-7 = 11\/30,\u00a0x\u00a0= -4, 7<\/h2>\n

Solution:
\n(i)\u00a0x-1\/x\u00a0= 3
\n\u21d2\u00a0x2\u00a0\u2013 3x\u00a0-1 = 0
\nOn comparing the given equation with\u00a0ax2\u00a0+\u00a0bx\u00a0+\u00a0c\u00a0= 0, we get
\na\u00a0= 1,\u00a0b\u00a0= -3 and\u00a0c\u00a0= -1
\nBy using quadratic formula, we get,<\/h3>\n

\"\"
\n\u21d2\u00a0x\u00a0= (3\u00b1\u221a9+4)\/2
\n\u21d2\u00a0x\u00a0= (3\u00b1\u221a13)\/2
\n\u2234\u00a0x\u00a0= (3+\u221a13)\/2 or\u00a0x\u00a0= (3-\u221a13)\/2<\/h3>\n

(ii) 1\/x+4 \u2013 1\/x-7 = 11\/30<\/h3>\n

\u21d2\u00a0x-7-x-4\/(x+4)(x-7) = 11\/30
\n\u21d2 -11\/(x+4)(x-7) = 11\/30
\n\u21d2 (x+4)(x-7) = -30
\n\u21d2\u00a0x2\u00a0\u2013 3x\u00a0\u2013 28 = 30
\n\u21d2\u00a0x2\u00a0\u2013 3x\u00a0+ 2 = 0
\nWe can solve this equation by factorization method now,
\n\u21d2\u00a0x2\u00a0\u2013 2x\u00a0\u2013\u00a0x\u00a0+ 2 = 0
\n\u21d2\u00a0x(x\u00a0\u2013 2) \u2013 1(x\u00a0\u2013 2) = 0
\n\u21d2 (x\u00a0\u2013 2)(x\u00a0\u2013 1) = 0
\n\u21d2\u00a0x\u00a0= 1 or 2<\/h3>\n

4.\u00a0The sum of the reciprocals of Rehman\u2019s ages, (in years) 3 years ago and 5 years from now is 1\/3. Find his present age.<\/h2>\n

Solution:
\nLet us say, present age of Rahman is\u00a0x\u00a0years.
\nThree years ago, Rehman\u2019s age was (x\u00a0\u2013 3) years.
\nFive years after, his age will be (x\u00a0+ 5) years.
\nGiven, the sum of the reciprocals of Rehman\u2019s ages 3 years ago and after 5 years is equal to 1\/3.
\n\u2234 1\/x-3\u00a0+ 1\/x-5 = 1\/3
\n(x+5+x-3)\/(x-3)(x+5) = 1\/3
\n(2x+2)\/(x-3)(x+5) = 1\/3
\n\u21d2 3(2x\u00a0+\u00a02) = (x-3)(x+5)
\n\u21d2 6x\u00a0+ 6 =\u00a0x2\u00a0+ 2x\u00a0\u2013 15
\n\u21d2\u00a0x2\u00a0\u2013\u00a04x\u00a0\u2013 21 = 0
\n\u21d2\u00a0x2\u00a0\u2013 7x\u00a0+ 3x\u00a0\u2013 21 = 0
\n\u21d2\u00a0x(x\u00a0\u2013 7)\u00a0+ 3(x\u00a0\u2013 7) = 0
\n\u21d2 (x\u00a0\u2013 7)(x\u00a0+ 3) = 0
\n\u21d2\u00a0x\u00a0= 7, -3
\nAs we know, age cannot be negative.
\nTherefore, Rahman\u2019s present age is 7 years.<\/h3>\n

5. In a class test, the sum of Shefali\u2019s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.<\/h2>\n

Solution:
\nLet us say, the marks of Shefali in Maths be\u00a0x.
\nThen, the marks in English will be 30 \u2013\u00a0x.
\nAs per the given question,
\n(x\u00a0+ 2)(30 \u2013\u00a0x\u00a0\u2013 3) = 210
\n(x\u00a0+ 2)(27 \u2013\u00a0x) = 210
\n\u21d2 -x2\u00a0+ 25x\u00a0+ 54 = 210
\n\u21d2\u00a0x2\u00a0\u2013 25x\u00a0+ 156 = 0
\n\u21d2\u00a0x2\u00a0\u2013\u00a012x\u00a0\u2013 13x\u00a0+ 156 = 0
\n\u21d2\u00a0x(x\u00a0\u2013 12) -13(x\u00a0\u2013 12) = 0
\n\u21d2 (x\u00a0\u2013 12)(x\u00a0\u2013 13) = 0
\n\u21d2\u00a0x\u00a0= 12, 13
\nTherefore, if the marks in Maths are 12, then marks in English will be 30 \u2013 12 = 18 and the marks in Maths are 13, then marks in English will be 30 \u2013 13 = 17.<\/h3>\n

6.\u00a0The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.<\/h2>\n

Solution:
\nLet us say, the shorter side of the rectangle be\u00a0x\u00a0m.
\nThen, larger side of the rectangle = (x\u00a0+ 30) m<\/h3>\n

\"\"
\nAs given, the length of the diagonal is = x + 30 m
\nTherefore,<\/h3>\n

\"\"
\n\u21d2\u00a0x2\u00a0+ (x\u00a0+ 30)2\u00a0= (x\u00a0+ 60)2
\n\u21d2\u00a0x2\u00a0+\u00a0x2\u00a0+ 900\u00a0+ 60x\u00a0=\u00a0x2\u00a0+ 3600\u00a0+ 120x
\n\u21d2\u00a0x2\u00a0\u2013 60x\u00a0\u2013 2700 = 0
\n\u21d2\u00a0x2\u00a0\u2013 90x\u00a0+ 30x\u00a0\u2013 2700 = 0
\n\u21d2\u00a0x(x\u00a0\u2013 90)\u00a0+ 30(x\u00a0-90) = 0
\n\u21d2 (x\u00a0\u2013 90)(x\u00a0+ 30) = 0
\n\u21d2\u00a0x\u00a0= 90, -30
\nHowever, side of the field cannot be negative. Therefore, the length of the shorter side will be\u00a090 m.
\nand the length of the larger side will be (90 + 30) m = 120 m.<\/h3>\n

7.\u00a0The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.<\/h2>\n

Solution:
\nLet us say, the larger and smaller number be\u00a0x\u00a0and\u00a0y\u00a0respectively.
\nAs per the question given,
\nx2\u00a0\u2013\u00a0y2\u00a0= 180 and\u00a0y2\u00a0= 8x
\n\u21d2\u00a0x2\u00a0\u2013 8x\u00a0= 180
\n\u21d2\u00a0x2\u00a0\u2013\u00a08x\u00a0\u2013 180 = 0
\n\u21d2\u00a0x2\u00a0\u2013 18x\u00a0+ 10x\u00a0\u2013 180 = 0
\n\u21d2\u00a0x(x\u00a0\u2013 18)\u00a0+10(x\u00a0\u2013 18) = 0
\n\u21d2 (x\u00a0\u2013 18)(x\u00a0+ 10) = 0
\n\u21d2\u00a0x\u00a0= 18, -10
\nHowever, the larger number cannot considered as negative number, as 8 times of the larger number will be negative and hence, the square of the smaller number will be negative which is not possible.
\nTherefore, the larger number will be 18 only.
\nx\u00a0= 18
\n\u2234\u00a0y2\u00a0= 8x = 8 \u00d7 18 = 144
\n\u21d2\u00a0y\u00a0= \u00b1\u221a144\u00a0= \u00b112
\n\u2234 Smaller number = \u00b112
\nTherefore, the numbers are 18 and 12 or 18 and -12.<\/h3>\n

8. A train travels 360 km at a uniform speed. If the speed had been 5 km\/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.<\/h2>\n

Solution:
\nLet us say, the speed of the train be\u00a0x\u00a0km\/hr.
\nTime taken to cover 360 km = 360\/x\u00a0hr.
\nAs per the question given,
\n\u21d2 (x\u00a0+ 5)(360\/x \u2013 1)\u00a0= 360
\n\u21d2 360 \u2013\u00a0x\u00a0+ 1800\/x-5\u00a0= 360
\n\u21d2\u00a0x2\u00a0+\u00a05x\u00a0\u2013 1800 = 0
\n\u21d2\u00a0x2\u00a0+ 45x \u2013 40x \u2013 1800 = 0
\n\u21d2\u00a0x(x\u00a0+ 45) -40(x\u00a0+ 45) = 0
\n\u21d2 (x\u00a0+ 45)(x\u00a0\u2013 40) = 0
\n\u21d2\u00a0x\u00a0= 40, -45
\nAs we know, the value of speed cannot be negative.
\nTherefore, the speed of train is 40 km\/h.<\/h3>\n

9.\u00a0Two water taps together can fill a tank in<\/h2>\n

\"\"<\/p>\n

hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.<\/h2>\n

Solution:<\/h3>\n

Let the time taken by the smaller pipe to fill the tank = x hr.<\/h3>\n

Time taken by the larger pipe = (x\u00a0\u2013 10) hr<\/h3>\n

Part of tank filled by smaller pipe in 1 hour = 1\/x<\/h3>\n

Part of tank filled by larger pipe in 1 hour = 1\/(x\u00a0\u2013 10)<\/h3>\n

As given, the tank can be filled in<\/h3>\n

\"\"
\n= 75\/8\u00a0hours by both the pipes together.
\nTherefore,
\n1\/x\u00a0+ 1\/x-10 = 8\/75
\nx-10+x\/x(x-10) = 8\/75
\n\u21d2 2x-10\/x(x-10) = 8\/75
\n\u21d2 75(2x\u00a0\u2013 10) = 8×2\u00a0\u2013 80x
\n\u21d2 150x\u00a0\u2013 750 = 8×2\u00a0\u2013 80x
\n\u21d2 8×2\u00a0\u2013 230x\u00a0+750 = 0
\n\u21d2 8×2\u00a0\u2013 200x\u00a0\u2013 30x\u00a0+ 750 = 0
\n\u21d2 8x(x\u00a0\u2013 25) -30(x\u00a0\u2013 25) = 0
\n\u21d2 (x\u00a0\u2013 25)(8x\u00a0-30) = 0
\n\u21d2\u00a0x\u00a0= 25, 30\/8
\nTime taken by the smaller pipe cannot be 30\/8\u00a0= 3.75 hours, as the time taken by the larger pipe will become negative, which is logically not possible.
\nTherefore, time taken individually by the smaller pipe and the larger pipe will be 25 and 25 \u2013 10 =15 hours respectively.<\/h3>\n

10. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speeds of the express train is 11 km\/h more than that of the passenger train, find the average speed of the two trains.<\/h2>\n

Solution:
\nLet us say, the average speed of passenger train = \u00a0x\u00a0km\/h.
\nAverage speed of express train = (x\u00a0+ 11) km\/h
\nGiven, time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance. Therefore,
\n(132\/x) \u2013 (132\/(x+11)) = 1
\n132(x+11-x)\/(x(x+11)) = 1
\n132 \u00d7 11 \/(x(x+11)) = 1
\n\u21d2 132 \u00d7 11 =\u00a0x(x\u00a0+ 11)
\n\u21d2\u00a0x2\u00a0+ 11x\u00a0\u2013 1452 = 0
\n\u21d2\u00a0x2\u00a0+\u00a0 44x\u00a0-33x\u00a0-1452 = 0
\n\u21d2\u00a0x(x\u00a0+ 44) -33(x\u00a0+ 44) = 0
\n\u21d2 (x\u00a0+ 44)(x\u00a0\u2013 33) = 0
\n\u21d2\u00a0x\u00a0= \u2013 44, 33
\nAs we know, Speed cannot be negative.
\nTherefore, the speed of the passenger train will be 33 km\/h and thus, the speed of the express train will be 33 + 11 = 44 km\/h.<\/h3>\n

11. Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.<\/h2>\n

Solution:
\nLet the sides of the two squares be\u00a0x\u00a0m and\u00a0y\u00a0m.
\nTherefore, their perimeter will be 4x\u00a0and 4y\u00a0respectively
\nAnd area of the squares will be\u00a0x2\u00a0and\u00a0y2\u00a0respectively.
\nGiven,
\n4x\u00a0\u2013 4y\u00a0= 24
\nx\u00a0\u2013\u00a0y\u00a0= 6
\nx\u00a0=\u00a0y\u00a0+ 6
\nAlso,\u00a0x2\u00a0+\u00a0y2\u00a0= 468
\n\u21d2 (6\u00a0+\u00a0y2)\u00a0+\u00a0y2\u00a0= 468
\n\u21d2 36\u00a0+\u00a0y2\u00a0+ 12y\u00a0+\u00a0y2\u00a0= 468
\n\u21d2 2y2\u00a0+ 12y\u00a0+ 432 = 0
\n\u21d2\u00a0y2\u00a0+ 6y \u2013 216 = 0
\n\u21d2\u00a0y2\u00a0+ 18y\u00a0\u2013 12y\u00a0\u2013 216 = 0
\n\u21d2\u00a0y(y\u00a0+18) -12(y\u00a0+ 18) = 0
\n\u21d2 (y\u00a0+ 18)(y\u00a0\u2013 12) = 0
\n\u21d2\u00a0y\u00a0= -18, 12
\nAs we know, the side of a square cannot be negative.
\nHence, the sides of the squares are 12 m and (12 + 6) m = 18 m.<\/h3>\n

Exercise 4.4<\/h2>\n

1. Find the nature of the roots of the following quadratic equations.\u00a0If the real roots exist, find them;
\n(i) 2×2\u00a0\u2013 3x\u00a0+ 5 = 0
\n(ii) 3×2\u00a0\u2013\u00a04\u221a3x\u00a0+ 4 = 0
\n(iii) 2×2\u00a0\u2013\u00a06x\u00a0+ 3 = 0<\/h2>\n

Solutions:
\n(i) Given,
\n2×2\u00a0\u2013 3x\u00a0+ 5 = 0
\nComparing the equation with\u00a0ax2\u00a0+\u00a0bx\u00a0+\u00a0c\u00a0= 0, we get
\na\u00a0= 2,\u00a0b\u00a0= -3 and\u00a0c\u00a0= 5
\nWe know, Discriminant =\u00a0b2\u00a0\u2013 4ac
\n=\u00a0( \u2013 3)2\u00a0\u2013 4 (2) (5) = 9 \u2013 40
\n= \u2013 31
\nAs you can see, b2\u00a0\u2013 4ac < 0 Therefore, no real root is possible for the given equation,\u00a02×2\u00a0\u2013 3x\u00a0+ 5 = 0. (ii) 3×2\u00a0\u2013 4\u221a3x\u00a0+ 4 = 0 Comparing the equation with\u00a0ax2\u00a0+\u00a0bx\u00a0+\u00a0c\u00a0= 0, we get a\u00a0= 3,\u00a0b\u00a0=\u00a0-4\u221a3\u00a0and\u00a0c\u00a0= 4 We know, Discriminant =\u00a0b2\u00a0\u2013 4ac = (-4\u221a3)2\u00a0\u2013 4(3)(4) = 48 \u2013 48 = 0 As\u00a0b2\u00a0\u2013 4ac\u00a0= 0, Real roots exist for the given equation and they are equal to each other. Hence the roots will be \u2013b\/2a\u00a0and\u00a0\u2013b\/2a. \u2013b\/2a\u00a0= -(-4\u221a3)\/2\u00d73 = 4\u221a3\/6 = 2\u221a3\/3 = 2\/\u221a3 Therefore, the roots are\u00a02\/\u221a3\u00a0and 2\/\u221a3. (iii) 2×2\u00a0\u2013\u00a06x\u00a0+ 3 = 0 Comparing the equation with\u00a0ax2\u00a0+\u00a0bx\u00a0+\u00a0c\u00a0= 0, we get a\u00a0= 2,\u00a0b\u00a0= -6,\u00a0c\u00a0= 3 As we know, Discriminant =\u00a0b2\u00a0\u2013 4ac = (-6)2\u00a0\u2013 4 (2) (3) = 36 \u2013 24 = 12 As\u00a0b2\u00a0\u2013 4ac\u00a0> 0,
\nTherefore, there are distinct real roots exist for this equation, 2×2\u00a0\u2013\u00a06x\u00a0+ 3 = 0.<\/h3>\n

\"\"
\n= (-(-6) \u00b1 \u221a(-62-4(2)(3)) )\/ 2(2)
\n= (6\u00b12\u221a3 )\/4
\n= (3\u00b1\u221a3)\/2
\nTherefore the roots for the given equation are (3+\u221a3)\/2 and (3-\u221a3)\/2<\/h3>\n

2. Find the values of\u00a0k\u00a0for each of the following quadratic equations, so that they have two equal roots.
\n(i) 2×2\u00a0+\u00a0kx\u00a0+ 3 = 0
\n(ii)\u00a0kx\u00a0(x\u00a0\u2013 2) + 6 = 0<\/h2>\n

Solutions:
\n(i) 2×2\u00a0+\u00a0kx\u00a0+ 3 = 0
\nComparing the given equation with\u00a0ax2\u00a0+\u00a0bx\u00a0+\u00a0c\u00a0= 0, we get,
\na\u00a0= 2,\u00a0b\u00a0= k and\u00a0c\u00a0= 3
\nAs we know, Discriminant =\u00a0b2\u00a0\u2013 4ac
\n= (k)2\u00a0\u2013 4(2) (3)
\n=\u00a0k2\u00a0\u2013 24
\nFor equal roots, we know,
\nDiscriminant = 0
\nk2\u00a0\u2013 24 = 0
\nk2\u00a0= 24
\nk = \u00b1\u221a24\u00a0= \u00b12\u221a6<\/h3>\n

(ii)\u00a0kx(x\u00a0\u2013 2) + 6 = 0
\nor\u00a0kx2\u00a0\u2013 2kx\u00a0+ 6 = 0
\nComparing the given equation with\u00a0ax2\u00a0+\u00a0bx\u00a0+\u00a0c\u00a0= 0, we get
\na\u00a0=\u00a0k,\u00a0b\u00a0= \u2013 2k\u00a0and\u00a0c\u00a0= 6
\nWe know, Discriminant =\u00a0b2\u00a0\u2013 4ac
\n= ( \u2013 2k)2\u00a0\u2013 4 (k) (6)
\n= 4k2\u00a0\u2013 24k
\nFor equal roots, we know,
\nb2\u00a0\u2013 4ac\u00a0= 0
\n4k2\u00a0\u2013 24k\u00a0= 0
\n4k\u00a0(k\u00a0\u2013 6) = 0
\nEither 4k\u00a0= 0 or\u00a0k\u00a0= 6 = 0
\nk\u00a0= 0 or\u00a0k\u00a0= 6
\nHowever, if\u00a0k\u00a0= 0, then the equation will not have the terms \u2018x2\u2018 and \u2018x\u2018.
\nTherefore, if this equation has two equal roots,\u00a0k\u00a0should be 6 only.<\/h3>\n

3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2?\u00a0If so, find its length and breadth.<\/h2>\n

Solution:
\nLet the breadth of mango grove be\u00a0l.
\nLength of mango grove will be 2l.
\nArea of mango grove = (2l) (l)= 2l2
\n2l2\u00a0= 800
\nl2\u00a0= 800\/2 = 400
\nl2\u00a0\u2013 400 =0
\nComparing the given equation with\u00a0ax2\u00a0+\u00a0bx\u00a0+\u00a0c\u00a0= 0, we get
\na\u00a0= 1,\u00a0b\u00a0= 0,\u00a0c\u00a0= 400
\nAs we know, Discriminant =\u00a0b2\u00a0\u2013 4ac
\n=> (0)2\u00a0\u2013 4 \u00d7 (1) \u00d7 ( \u2013 400) = 1600
\nHere,\u00a0b2\u00a0\u2013 4ac\u00a0> 0
\nThus, the equation will have real roots. And hence, the desired rectangular mango grove can be designed.
\nl\u00a0= \u00b120
\nAs we know, the value of length cannot be negative.
\nTherefore, breadth of mango grove = 20 m
\nLength of mango grove = 2 \u00d7 20 = 40 m<\/h3>\n

4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.<\/h2>\n

Solution:
\nLet\u2019s say, the age of one friend be x years.
\nThen, the age of the other friend will be (20 \u2013 x) years.
\nFour years ago,
\nAge of First friend = (x\u00a0\u2013 4) years
\nAge of Second friend = (20 \u2013\u00a0x\u00a0\u2013 4)\u00a0= (16 \u2013\u00a0x) years
\nAs per the given question, we can write,
\n(x\u00a0\u2013 4) (16 \u2013\u00a0x) = 48
\n16x \u2013 x2\u00a0\u2013 64 + 4x\u00a0= 48
\n\u2013 x2\u00a0+\u00a020x \u2013\u00a0112 = 0
\nx2\u00a0\u2013\u00a020x +\u00a0112 = 0
\nComparing the equation with\u00a0ax2\u00a0+\u00a0bx\u00a0+\u00a0c\u00a0= 0, we get
\na\u00a0=\u00a01,\u00a0b\u00a0= -20\u00a0and\u00a0c\u00a0= 112
\nDiscriminant =\u00a0b2\u00a0\u2013 4ac
\n= (-20)2\u00a0\u2013 4 \u00d7 112
\n= 400 \u2013 448 = -48
\nb2\u00a0\u2013 4ac\u00a0< 0
\nTherefore, there will be no real solution possible for the equations. Hence, condition doesn\u2019t exist.<\/h3>\n

5.\u00a0Is it possible to design a rectangular park of perimeter 80 and area 400 m2? If so find its length and breadth.<\/h2>\n

Solution:
\nLet the length and breadth of the park be\u00a0l\u00a0and b.
\nPerimeter of the rectangular park = 2 (l + b) = 80
\nSo, l + b\u00a0= 40
\nOr,\u00a0b\u00a0= 40 \u2013\u00a0l
\nArea of the rectangular park =\u00a0l\u00d7b = l(40 \u2013 l) =\u00a040l\u00a0\u2013\u00a0l2\u00a0= 400
\nl2\u00a0\u2013\u00a0\u00a040l\u00a0+ 400\u00a0= 0, which is a quadratic equation.
\nComparing the equation with\u00a0ax2\u00a0+\u00a0bx\u00a0+\u00a0c\u00a0= 0, we get
\na\u00a0= 1,\u00a0b\u00a0= -40,\u00a0c\u00a0= 400
\nSince, Discriminant =\u00a0b2\u00a0\u2013 4ac
\n=(-40)2\u00a0\u2013 4 \u00d7 400
\n= 1600 \u2013 1600 = 0
\nThus,\u00a0b2\u00a0\u2013 4ac\u00a0= 0
\nTherefore, this equation has equal real roots. Hence, the situation is possible.
\nRoot of the equation,
\nl\u00a0= \u2013b\/2a
\nl\u00a0= -(-40)\/2(1) = 40\/2 = 20
\nTherefore, length of rectangular park,\u00a0l\u00a0= 20 m
\nAnd breadth of the park,\u00a0b\u00a0= 40 \u2013\u00a0l\u00a0= 40 \u2013 20 = 20 m.<\/h3>\n","protected":false},"excerpt":{"rendered":"

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