{"id":121815,"date":"2022-05-19T11:41:47","date_gmt":"2022-05-19T06:11:47","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=121815"},"modified":"2022-05-19T11:41:47","modified_gmt":"2022-05-19T06:11:47","slug":"chapter-5-complex-numbers-and-quadratic-equations-questions-and-answers-ncert-solutions-for-class-11-maths","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/education\/chapter-5-complex-numbers-and-quadratic-equations-questions-and-answers-ncert-solutions-for-class-11-maths","title":{"rendered":"Chapter 5 – Complex Numbers and Quadratic Equations Questions and Answers: NCERT Solutions for Class 11 Maths"},"content":{"rendered":"

Exercise 5.1<\/h2>\n

1. Express the given complex number in the form a+ib:(5i)(\u221235i)a+ib:(5i)(\u221235i)
\nAnd evaluate<\/h2>\n

Ans:
\nEvaluate the complex number
\n(5i)(\u221235i)=\u22125\u00d735\u00d7i\u00d7i(5i)(\u221235i)=\u22125\u00d735\u00d7i\u00d7i
\n(5i)(\u221235i)=\u22123i2\u22ef[i2=\u22121](5i)(\u221235i)=\u22123i2\u22ef[i2=\u22121]
\n(5i)(\u221235i)=3(5i)(\u221235i)=3<\/h3>\n

2. Express the given complex number in the form a+ib:i9+i19a+ib:i9+i19
\nAnd evaluate<\/h2>\n

Ans:
\nEvaluate the complex number
\ni9+i19=i4\u00d72+1+i4\u00d74+3i9+i19=i4\u00d72+1+i4\u00d74+3
\ni9+i19=(i4)2.i+(i4)4.i3\u22ef[i4=1,i3=\u22121]i9+i19=(i4)2.i+(i4)4.i3\u22ef[i4=1,i3=\u22121]
\ni9+i19=0i9+i19=0<\/h3>\n

3. Express the given complex number in the form a+ib:i\u221239a+ib:i\u221239
\nAnd evaluate<\/h2>\n

Ans:
\nEvaluate the complex number
\ni\u221239=i4\u00d79\u22123i\u221239=i4\u00d79\u22123
\ni\u221239=(i4)\u22129.i\u22123i\u221239=(i4)\u22129.i\u22123
\ni\u221239=i\u22ef[i=\u22121]i\u221239=i\u22ef[i=\u22121]
\ni\u221239=ii\u221239=i<\/h3>\n

4. Express the given complex number in the form a+ib:3(7+i7)+i(7+i7)a+ib:3(7+i7)+i(7+i7)
\nAnd evaluate<\/h2>\n

Ans:
\nEvaluate the complex number
\n3(7+i7)+i(7+i7)=21+21i+7i+7i23(7+i7)+i(7+i7)=21+21i+7i+7i2
\n3(7+i7)+i(7+i7)=21+28i+7i2\u22ef[i2=\u22121]3(7+i7)+i(7+i7)=21+28i+7i2\u22ef[i2=\u22121]
\n3(7+i7)+i(7+i7)=14+28i3(7+i7)+i(7+i7)=14+28i<\/h3>\n

5. Express the given complex number in the form
\na+ib:(1\u2212i)\u2212(\u22121+6i)a+ib:(1\u2212i)\u2212(\u22121+6i)
\nAnd evaluate<\/h2>\n

Ans:
\nEvaluate the complex number
\n(1\u2212i)\u2212(\u22121+6i)=1\u2212i+1\u2212i6(1\u2212i)\u2212(\u22121+6i)=1\u2212i+1\u2212i6
\n(1\u2212i)\u2212(\u22121+6i)=2\u22127i(1\u2212i)\u2212(\u22121+6i)=2\u22127i<\/h3>\n

6. Express the given complex number in the form a+ib:(15+i25)\u2212(4+i52)a+ib:(15+i25)\u2212(4+i52)
\nAnd evaluate<\/h2>\n

Ans:
\nEvaluate the complex number (15+i25)\u2212(4+i52)=15+i25\u22124\u2212i52(15+i25)\u2212(4+i52)=15+i25\u22124\u2212i52
\n(15+i25)\u2212(4+i52)=\u2212195+i[\u22122110](15+i25)\u2212(4+i52)=\u2212195+i[\u22122110]
\n(15+i25)\u2212(4+i52)=\u2212195\u22122110i(15+i25)\u2212(4+i52)=\u2212195\u22122110i<\/h3>\n

7. Express the given complex number in the form a+ib:[(13+i73)+(4+i13)\u2212(\u221243+i)]a+ib:[(13+i73)+(4+i13)\u2212(\u221243+i)]
\nAnd evaluate<\/h2>\n

Ans:
\nEvaluate the complex number
\n[(13+i73)+(4+i13)\u2212(\u221243+i)]=13+i73+4+i13+43\u2212i[(13+i73)+(4+i13)\u2212(\u221243+i)]=13+i73+4+i13+43\u2212i
\n[(13+i73)+(4+i13)\u2212(\u221243+i)]=(13+4+43)+i(73+13\u22121)[(13+i73)+(4+i13)\u2212(\u221243+i)]=(13+4+43)+i(73+13\u22121)
\n[(13+i73)+(4+i13)\u2212(\u221243+i)]=173+i53[(13+i73)+(4+i13)\u2212(\u221243+i)]=173+i53<\/h3>\n

8. Express the given complex number in the form a+ib:(1\u2212i)4a+ib:(1\u2212i)4
\nAnd evaluate<\/h2>\n

Ans:
\nEvaluate the complex number
\n(1\u2212i)4=[1+i2\u22122i]2(1\u2212i)4=[1+i2\u22122i]2
\n(1\u2212i)4=[1\u22121\u22122i]2(1\u2212i)4=[1\u22121\u22122i]2
\n(1\u2212i)4=(\u22122i)\u00d7(\u22122i)(1\u2212i)4=(\u22122i)\u00d7(\u22122i)
\n(1\u2212i)4=\u22124(1\u2212i)4=\u22124<\/h3>\n

9. Express the given complex number in the form a+ib:(13+3i)3a+ib:(13+3i)3
\nAnd evaluate<\/h2>\n

Ans:
\nEvaluate the complex number
\n(13+3i)3=(13)3+(3i)3+333i(13+3i)(13+3i)3=(13)3+(3i)3+333i(13+3i)
\n(13+3i)3=127\u2212(27i)+3i(13+3i)(13+3i)3=127\u2212(27i)+3i(13+3i)
\n(13+3i)3=\u221224227\u221226i(13+3i)3=\u221224227\u221226i<\/h3>\n

10. Express the given complex number in the form a+ib:(\u22122\u221213i)3a+ib:(\u22122\u221213i)3
\nAnd evaluate<\/h2>\n

Ans:
\nEvaluate the complex number
\n(\u22122\u221213i)3=(\u22121)3(2+13i)3(\u22122\u221213i)3=(\u22121)3(2+13i)3
\n(\u22122\u221213i)3=\u2212(23+(i3)3+6i3(2+i3))(\u22122\u221213i)3=\u2212(23+(i3)3+6i3(2+i3))
\n(\u22122\u221213i)3=\u2212223\u221210727i(\u22122\u221213i)3=\u2212223\u221210727i<\/h3>\n

11. Find the multiplicative inverse of the complex number
\n4\u22123i4\u22123i
\nAnd evaluate<\/h2>\n

Ans:
\nLet z=4\u22123iz=4\u22123i
\nThen,
\nz\u00af\u00af\u00af=4+3i&|z\u00af\u00af\u00af|=42+(\u22123)2=16+9=25z\u00af=4+3i&|z\u00af|=42+(\u22123)2=16+9=25
\nTherefore, the multiplicative inverse of 4\u22123i4\u22123i is given by
\nz\u22121=z\u00af\u00af\u00af|z|2=4+3i25=425+325iz\u22121=z\u00af|z|2=4+3i25=425+325i
\nHere we got final answer<\/h3>\n

12. Find the multiplicative inverse of the complex number 5\u2013\u221a+3i5+3i
\nAnd evaluate<\/h2>\n

Ans:
\nLet z=5\u2013\u221a+3iz=5+3i
\nThen,
\nz\u00af=5\u2013\u221a\u22123iand|z|2=(5\u2013\u221a)2+32=5+9=14z\u00af=5\u22123iand|z|2=(5)2+32=5+9=14
\nTherefore, the multiplicative inverse of 5\u2013\u221a+3i5+3i is given by
\nz\u22121=z\u00af|z|2=5\u2013\u221a\u22123i14=5\u2013\u221a14\u22123i14z\u22121=z\u00af|z|2=5\u22123i14=514\u22123i14
\nHere we got final answer<\/h3>\n

13. Find the multiplicative inverse of the complex number
\n\u2212i\u2212i
\nAnd evaluate<\/h2>\n

Ans:
\nLet z=\u2212iz=\u2212i
\nThen,
\nz\u00af=iand|z|2=12=1z\u00af=iand|z|2=12=1
\nTherefore, the multiplicative inverse of \u2212i\u2212i is given by z\u22121=z\u00af|z|2=i1=iz\u22121=z\u00af|z|2=i1=i
\nHere we got final answer<\/h3>\n

14. Express the following expression in the form of a+iba+ib (3+i5\u2013\u221a)(3\u2212i5\u2013\u221a)(3\u2013\u221a+i2\u2013\u221a)\u2212(3\u2013\u221a\u2212i2\u2013\u221a)(3+i5)(3\u2212i5)(3+i2)\u2212(3\u2212i2)
\nEvaluate<\/h2>\n

Ans:
\nThe following expression (3+i5\u2013\u221a)(3\u2212i5\u2013\u221a)(3\u2013\u221a+2\u2013\u221ai)\u2212(3\u2013\u221a\u2212i2\u2013\u221a)=(3)2\u2212(i5\u2013\u221a)23\u2013\u221a+2\u2013\u221ai\u22123\u2013\u221a+2\u2013\u221ai(3+i5)(3\u2212i5)(3+2i)\u2212(3\u2212i2)=(3)2\u2212(i5)23+2i\u22123+2i
\n(3+i5\u2013\u221a)(3\u2212i5\u2013\u221a)(3\u2013\u221a+2\u2013\u221ai)\u2212(3\u2013\u221a\u2212i2\u2013\u221a)=9\u22125i222\u2013\u221ai(3+i5\u2013\u221a)(3\u2212i5\u2013\u221a)(3\u2013\u221a+2\u2013\u221ai)\u2212(3\u2013\u221a\u2212i2\u2013\u221a)=9\u22125(\u22121)22\u2013\u221ai(3+i5\u2013\u221a)(3\u2212i5\u2013\u221a)(3\u2013\u221a+2\u2013\u221ai)\u2212(3\u2013\u221a\u2212i2\u2013\u221a)=\u221272i\u2212\u2212\u221a2(3+i5)(3\u2212i5)(3+2i)\u2212(3\u2212i2)=9\u22125i222i(3+i5)(3\u2212i5)(3+2i)\u2212(3\u2212i2)=9\u22125(\u22121)22i(3+i5)(3\u2212i5)(3+2i)\u2212(3\u2212i2)=\u221272i2
\nHere we got final answer<\/h3>\n

Exercise 5.2<\/h2>\n

1. Find the modulus and the argument of the complex number z=\u22121\u2212i3\u2013\u221az=\u22121\u2212i3
\nEvaluate<\/h2>\n

Ans:
\nThe complex number is
\nz=\u22121\u2212i3\u2013\u221az=\u22121\u2212i3
\nLet rcos\u03b8=-1andrsin\u03b8=-3\u2013\u221arcos\u03b8=-1andrsin\u03b8=-3
\nSquaring and adding
\n(rcos\u03b8)2+(rsin\u03b8)2=(\u22121)2+(\u22123\u2013\u221a)2(rcos\u03b8)2+(rsin\u03b8)2=(\u22121)2+(\u22123)2
\nr2(cos2\u03b8+sin2\u03b8)=1+3r2=4[cos2\u03b8+sin2\u03b8=1]r2(cos2\u03b8+sin2\u03b8)=1+3r2=4[cos2\u03b8+sin2\u03b8=1]
\nr=4\u2013\u221a=2[Conventionally,r0]r=4=2[Conventionally,r0]
\nModulus=22cos\u03b8=-1and2sin\u03b8=-3\u2013\u221acos\u03b8=\u221212andsin\u03b8=\u22123\u2013\u221a2Modulus=22cos\u03b8=-1and2sin\u03b8=-3cos\u03b8=\u221212andsin\u03b8=\u221232
\nSince both the values of sin\u03b8andcos\u03b8sin\u03b8andcos\u03b8 negative and sin\u03b8andcos\u03b8sin\u03b8andcos\u03b8 are negative in 3rd quadrant,
\nArgument=-(\u03c0-\u03c03)=\u22122\u03c03Argument=-(\u03c0-\u03c03)=\u22122\u03c03
\nThus, the modulus and argument of the complex number \u22121\u22123\u2013\u221aiare2and-2\u03c03\u22121\u22123iare2and-2\u03c03
\nRespectively<\/h3>\n

2. Find the modulus and the argument of the complex number
\nz=\u22123\u2013\u221a+iz=\u22123+i
\nEvaluate<\/h2>\n

Ans:
\nThe complex number is
\nz=\u22123\u2013\u221a+iz=\u22123+i
\nLet rcos\u03b8=-3\u2013\u221aandrsin\u03b8=1rcos\u03b8=-3andrsin\u03b8=1
\nsquaring and adding
\n(rcos\u03b8)2+(rsin\u03b8)2=(\u22123\u2013\u221a)2+(\u22121)2(rcos\u03b8)2+(rsin\u03b8)2=(\u22123)2+(\u22121)2
\nr2=3+1=4LLL[cos2\u03b8+sin2\u03b8=1]r=4\u2013\u221a=2LLL[Conventionally,r0]r2=3+1=4LLL[cos2\u03b8+sin2\u03b8=1]r=4=2LLL[Conventionally,r0]
\nModulus=22cos\u03b8=-3\u2013\u221aand2sin\u03b8=1Modulus=22cos\u03b8=-3and2sin\u03b8=1
\ncos\u03b8=\u22123\u2013\u221a2andsin\u03b8=12\u03b8=\u03c0-\u03c06=5\u03c06LL[As\u03b8lies in the II quadrant]cos\u03b8=\u221232andsin\u03b8=12\u03b8=\u03c0-\u03c06=5\u03c06LL[As\u03b8lies in the II quadrant]
\nThus, the modulus and argument of the complex number \u22123\u2013\u221a+iare2and5\u03c06\u22123+iare2and5\u03c06
\nRespectively<\/h3>\n

3. Convert the given complex number in polar form
\n1\u2212i1\u2212i
\nAnd evaluate<\/h2>\n

Ans:
\nThe complex number is
\n1\u2212i1\u2212i
\nLet rcos\u03b8=1andrsin\u03b8=\u22121rcos\u2061\u03b8=1andrsin\u2061\u03b8=\u22121
\nsquaring and adding
\nr2cos2\u03b8+r2sin2\u03b8=12+(\u22121)2\u21d2r2(cos2\u03b8+sin2\u03b8)=1+1r2cos2\u03b8+r2sin2\u03b8=12+(\u22121)2\u21d2r2(cos2\u03b8+sin2\u03b8)=1+1
\nr2=2r=2\u2013\u221a[Conventionally,r>0]r2=2r=2[Conventionally,r>0]
\n2\u2013\u221acos\u03b8=1and2\u2013\u221asin\u03b8=-1cos\u03b8=12\u2013\u221aandsin\u03b8=-12\u2013\u221a\u03b8=-\u03c04[As\u03b8liesin the IV quadrant]2cos\u03b8=1and2sin\u03b8=-1cos\u03b8=12andsin\u03b8=-12\u03b8=-\u03c04[As\u03b8liesin the IV quadrant]
\n1-i=rcos\u03b8+irsin\u03b8=2\u2013\u221acos(\u2212\u03c04)+i2\u2013\u221asin(\u2212\u03c04)=2\u2013\u221a[cos(\u2212\u03c04)+isin(\u2212\u03c04)]1-i=rcos\u03b8+irsin\u03b8=2cos(\u2212\u03c04)+i2sin(\u2212\u03c04)=2[cos(\u2212\u03c04)+isin(\u2212\u03c04)]
\nRequired polar form<\/h3>\n

4. Convert the given complex number in polar form
\n\u22121+i\u22121+i
\nAnd evaluate<\/h2>\n

Ans:
\nThe complex number is
\n\u22121+i\u22121+i
\nLet rcos\u03b8=-1andrsin\u03b8=1rcos\u03b8=-1andrsin\u03b8=1
\nSquaring and adding
\nr2cos2\u03b8+r2sin2\u03b8=(-1)2+12r2(cos2\u03b8+sin2\u03b8)=1+1r2=2r=2\u2013\u221ar2cos2\u03b8+r2sin2\u03b8=(-1)2+12r2(cos2\u03b8+sin2\u03b8)=1+1r2=2r=2
\n2\u2013\u221acos\u03b8=-1and2\u2013\u221asin\u03b8=12cos\u03b8=-1and2sin\u03b8=1
\ncos\u03b8=-12\u2013\u221aand2\u2013\u221asin\u03b8=1\u03b8=\u03c0-\u03c04=3\u03c04L[As\u03b8lies in the II quadrant]cos\u03b8=-12and2sin\u03b8=1\u03b8=\u03c0-\u03c04=3\u03c04L[As\u03b8lies in the II quadrant]
\nIt can be written,
\n-1+i=rcos\u03b8+irsin\u03b8=2\u2013\u221acos3\u03c04+i2\u2013\u221asin3\u03c04=2\u2013\u221a(cos3\u03c04+isin3\u03c04)-1+i=rcos\u03b8+irsin\u03b8=2cos3\u03c04+i2sin3\u03c04=2(cos3\u03c04+isin3\u03c04)
\nRequired polar form<\/h3>\n

5. Convert the given complex number in polar form \u22121\u2212i\u22121\u2212i
\nAnd evaluate<\/h2>\n

Ans:
\nThe complex number is
\n\u22121\u2212i\u22121\u2212i
\nLet rcos\u03b8=-1andrsin\u03b8=-1rcos\u03b8=-1andrsin\u03b8=-1
\nSquaring and adding
\nr2cos2\u03b8+r2sin2\u03b8=(-1)2+(\u22121)2r2(cos2\u03b8+sin2\u03b8)=1+1r2=2r=2\u2013\u221ar2cos2\u03b8+r2sin2\u03b8=(-1)2+(\u22121)2r2(cos2\u03b8+sin2\u03b8)=1+1r2=2r=2
\n2\u2013\u221acos\u03b8=-1and2\u2013\u221asin\u03b8=-1cos\u03b8=-12\u2013\u221aandsin\u03b8=-12\u2013\u221a\u03b8–(\u03c0-\u03c04)\u2212\u22123\u03c04[As0lies in the III quadrant]2cos\u03b8=-1and2sin\u03b8=-1cos\u03b8=-12andsin\u03b8=-12\u03b8–(\u03c0-\u03c04)\u2212\u22123\u03c04[As0lies in the III quadrant]
\n-1-i=rcos\u03b8+irsin\u03b8=2\u2013\u221acos\u22123\u03c04+i2\u2013\u221asin\u22123\u03c04=2\u2013\u221a(cos\u22123\u03c04+isin\u22123\u03c04)-1-i=rcos\u03b8+irsin\u03b8=2cos\u22123\u03c04+i2sin\u22123\u03c04=2(cos\u22123\u03c04+isin\u22123\u03c04)
\nRequired polar form<\/h3>\n

6. Convert the given complex number in polar form
\n\u22123\u22123
\nAnd evaluate<\/h2>\n

Ans:
\nThe complex number is
\n\u22123\u22123
\nLet rcos\u03b8=-3andrsin\u03b8=0rcos\u03b8=-3andrsin\u03b8=0
\nSquaring and adding
\nr2cos2\u03b8+r2sin2\u03b8=(-3)2r2(cos2\u03b8+sin2\u03b8)=9r2cos2\u03b8+r2sin2\u03b8=(-3)2r2(cos2\u03b8+sin2\u03b8)=9
\nr2=9r=9\u2013\u221a=3r2=9r=9=3
\n3cos\u03b8=-3and3sin\u03b8=0cos\u03b8=-1andsin=0\u03b8=\u03c03cos\u03b8=-3and3sin\u03b8=0cos\u03b8=-1andsin=0\u03b8=\u03c0
\n-3=rcos\u03b8+irsin\u03b8=3cos\u03c0+i3sin\u03c0=3(cos\u03c0+isin\u03c0)-3=rcos\u03b8+irsin\u03b8=3cos\u03c0+i3sin\u03c0=3(cos\u03c0+isin\u03c0)
\nRequired polar form<\/h3>\n

7. Convert the given complex number in polar form 3\u2013\u221a+i3+i
\nAnd evaluate<\/h2>\n

Ans:
\nThe complex number is
\n3\u2013\u221a+i3+i
\nLet rcos\u03b8=3\u2013\u221aandrsin\u03b8=1rcos\u03b8=3andrsin\u03b8=1
\nSquaring and adding
\nr2cos2\u03b8+r2sin2\u03b8=(3\u2013\u221a)2+12r2(cos2\u03b8+sin2\u03b8)=3+1r2=4r=4\u2013\u221a=2r2cos2\u03b8+r2sin2\u03b8=(3)2+12r2(cos2\u03b8+sin2\u03b8)=3+1r2=4r=4=2
\n2cos\u03b8=3\u2013\u221aand2sin\u03b8=1cos\u03b8=3\u2013\u221a2andsin\u03b8=12\u03b8=\u03c06[As\u03b8lies in the I quadrant]2cos\u03b8=3and2sin\u03b8=1cos\u03b8=32andsin\u03b8=12\u03b8=\u03c06[As\u03b8lies in the I quadrant]
\n3\u2013\u221a+i=rcos\u03b8+irsin\u03b8=2cos\u03c06+i2sin\u03c06=2(cos\u03c06+isin\u03c06)3+i=rcos\u03b8+irsin\u03b8=2cos\u03c06+i2sin\u03c06=2(cos\u03c06+isin\u03c06)
\nRequired polar form<\/h3>\n

8. Convert the given complex number in polar form
\nii
\nAnd evaluate<\/h2>\n

Ans:
\nThe complex number is
\nii
\nLet rcos\u03b8=0andrsin\u03b8=1rcos\u03b8=0andrsin\u03b8=1
\nSquaring and adding
\nr2cos2\u03b8+r2sin2\u03b8=02+12r2(cos2\u03b8+sin2\u03b8)=1r2cos2\u03b8+r2sin2\u03b8=02+12r2(cos2\u03b8+sin2\u03b8)=1
\nr2=1r=1\u2013\u221a=1[Conventionally,r0]r2=1r=1=1[Conventionally,r0]
\ncos\u03b8=0andsin\u03b8=1\u03b8=\u03c02i=rcos\u03b8+irsin\u03b8=cos\u03c02+isin\u03c02cos\u03b8=0andsin\u03b8=1\u03b8=\u03c02i=rcos\u03b8+irsin\u03b8=cos\u03c02+isin\u03c02
\nRequired polar form<\/h3>\n

Exercise 5.3<\/h2>\n

1. Solve the equation
\nx2+3=0x2+3=0
\nAnd evaluate<\/h2>\n

Ans:
\nQuadratic equation x2+3=0x2+3=0
\nGeneral form ax2+bx+c=0ax2+bx+c=0
\nWe obtain a=1,b=0,andc=3a=1,b=0,andc=3
\nTherefore, the discriminant of the given equation is
\nD=b2\u22124ac=02\u22124\u00d71\u00d73=-12D=b2\u22124ac=02\u22124\u00d71\u00d73=-12
\nTherefore, the required solutions are
\n=\u2212b\u00b1D\u2212\u2212\u221a2a=\u00b1\u221212\u2212\u2212\u2212\u2212\u221a2\u00d71=\u00b112\u2212\u2212\u221ai2=\u00b123\u2013\u221ai2=\u00b13\u2013\u221ai=\u2212b\u00b1D2a=\u00b1\u2212122\u00d71=\u00b112i2=\u00b123i2=\u00b13i<\/h3>\n

2. Solve the equation
\n2×2+x+1=02×2+x+1=0
\nAnd evaluate<\/h2>\n

Ans:
\nQuadratic equation 2×2+x+1=02×2+x+1=0
\nGeneral form ax2+bx+c=0ax2+bx+c=0
\nWe obtain a=2,b=1,andc=1a=2,b=1,andc=1
\nTherefore, the discriminant of the given equation is
\nD=b2\u22124ac=12\u22124\u00d72\u00d71=-7D=b2\u22124ac=12\u22124\u00d72\u00d71=-7
\nTherefore, the required solutions are
\n=\u2212b\u00b1D\u2212\u2212\u221a2a=\u00b1\u22127\u2212\u2212\u2212\u221a2\u00d72=\u00b17\u2013\u221ai4=\u2212b\u00b1D2a=\u00b1\u221272\u00d72=\u00b17i4<\/h3>\n

3. Solve the equation
\nx2+3x+9=0x2+3x+9=0
\nAnd evaluate<\/h2>\n

Ans:
\nQuadratic equation x2+3x+9=0x2+3x+9=0
\nGeneral form ax2+bx+c=0ax2+bx+c=0
\nWe obtain a=1,b=3,andc=9a=1,b=3,andc=9
\nTherefore, the discriminant of the given equation is
\nD=b2\u22124ac=32\u22124\u00d71\u00d79=-27D=b2\u22124ac=32\u22124\u00d71\u00d79=-27
\nTherefore, the required solutions are
\n=\u2212b\u00b1D\u2212\u2212\u221a2a=\u22123\u00b1\u221227\u2212\u2212\u2212\u2212\u221a2\u00d71=\u22123\u00b133\u2013\u221ai2=\u2212b\u00b1D2a=\u22123\u00b1\u2212272\u00d71=\u22123\u00b133i2<\/h3>\n

4. Solve the equation
\n\u2212x2+x\u22122=0\u2212x2+x\u22122=0
\nAnd evaluate<\/h2>\n

Ans:
\nQuadratic equation \u2212x2+x\u22122=0\u2212x2+x\u22122=0
\nGeneral form ax2+bx+c=0ax2+bx+c=0
\nWe obtain a=\u22121,b=1,andc=-2a=\u22121,b=1,andc=-2
\nTherefore, the discriminant of the given equation is
\nD=b2\u22124ac=12\u22124\u00d7-1\u00d7-2=-7D=b2\u22124ac=12\u22124\u00d7-1\u00d7-2=-7
\nTherefore, the required solutions are
\n=\u2212b\u00b1D\u2212\u2212\u221a2a=\u22121\u00b1\u22127\u2212\u2212\u2212\u221a2\u00d7-1=\u22121\u00b17\u2013\u221ai\u22122=\u2212b\u00b1D2a=\u22121\u00b1\u221272\u00d7-1=\u22121\u00b17i\u22122<\/h3>\n

5. Solve the equation
\nx2+3x+5=0x2+3x+5=0
\nAnd evaluate<\/h2>\n

Ans:
\nQuadratic equation x2+3x+5=0x2+3x+5=0
\nGeneral form ax2+bx+c=0ax2+bx+c=0
\nWe obtain a=1,b=3,andc=5a=1,b=3,andc=5
\nTherefore, the discriminant of the given equation is
\nD=b2\u22124ac=32\u22124\u00d71\u00d75=-11D=b2\u22124ac=32\u22124\u00d71\u00d75=-11
\nTherefore, the required solutions are
\n=\u2212b\u00b1D\u2212\u2212\u221a2a=\u22123\u00b1\u221211\u2212\u2212\u2212\u2212\u221a2\u00d71=\u22123\u00b111\u2212\u2212\u221ai2=\u2212b\u00b1D2a=\u22123\u00b1\u2212112\u00d71=\u22123\u00b111i2<\/h3>\n

6. Solve the equation
\nx2\u2212x+2=0x2\u2212x+2=0
\nAnd evaluate<\/h2>\n

Ans:
\nQuadratic equation x2\u2212x+2=0x2\u2212x+2=0
\nGeneral form ax2+bx+c=0ax2+bx+c=0
\nWe obtain a=1,b=3\u22121,andc=2a=1,b=3\u22121,andc=2
\nTherefore, the discriminant of the given equation is
\nD=b2\u22124ac=(\u22121)2\u22124\u00d71\u00d72=-7D=b2\u22124ac=(\u22121)2\u22124\u00d71\u00d72=-7
\nTherefore, the required solutions are
\n=\u2212b\u00b1D\u2212\u2212\u221a2a=\u2212(\u22121)\u00b1\u22127\u2212\u2212\u2212\u221a2\u00d71=1\u00b17\u2013\u221ai2=\u2212b\u00b1D2a=\u2212(\u22121)\u00b1\u221272\u00d71=1\u00b17i2<\/h3>\n

7. Solve the equation
\n2\u2013\u221ax2+x+2\u2013\u221a=02×2+x+2=0
\nAnd evaluate<\/h2>\n

Ans:
\nQuadratic equation 2\u2013\u221ax2+x+2\u2013\u221a=02×2+x+2=0
\nGeneral form ax2+bx+c=0ax2+bx+c=0
\nWe obtain a=2\u2013\u221a,b=1,andc=2\u2013\u221aa=2,b=1,andc=2
\nTherefore, the discriminant of the given equation is
\nD=b2\u22124ac=12\u22124\u00d72\u2013\u221a\u00d72\u2013\u221a=\u22127D=b2\u22124ac=12\u22124\u00d72\u00d72=\u22127
\nTherefore, the required solutions are
\n=\u2212b\u00b1D\u2212\u2212\u221a2a=\u22121\u00b1\u22127\u2212\u2212\u2212\u221a2\u00d72\u2013\u221a=\u22121\u00b17\u2013\u221ai22\u2013\u221a=\u2212b\u00b1D2a=\u22121\u00b1\u221272\u00d72=\u22121\u00b17i22<\/h3>\n

8. Solve the equation
\n3\u2013\u221ax2\u22122\u2013\u221ax+33\u2013\u221a=03×2\u22122x+33=0
\nAnd evaluate<\/h2>\n

Ans:
\nQuadratic equation 3\u2013\u221ax2\u22122\u2013\u221ax+33\u2013\u221a=03×2\u22122x+33=0
\nGeneral form ax2+bx+c=0ax2+bx+c=0
\nWe obtain a=3\u2013\u221a,b=\u22122\u2013\u221a,andc=33\u2013\u221aa=3,b=\u22122,andc=33
\nD=b2\u22124ac=(\u22122\u2013\u221a)2\u22124\u00d73\u2013\u221a\u00d733\u2013\u221a=\u221234D=b2\u22124ac=(\u22122)2\u22124\u00d73\u00d733=\u221234 D=b2\u22124ac=(\u22122\u2013\u221a)2\u22124\u00d73\u2013\u221a\u00d733\u2013\u221a=\u221234D=b2\u22124ac=(\u22122)2\u22124\u00d73\u00d733=\u221234
\nTherefore, the required solutions are
\n=\u2212b\u00b1D\u2212\u2212\u221a2a=\u2212(\u22122\u2013\u221a)\u00b1\u221234\u2212\u2212\u2212\u2212\u221a2\u00d73\u2013\u221a=2\u2013\u221a\u00b134\u2212\u2212\u221ai23\u2013\u221a=\u2212b\u00b1D2a=\u2212(\u22122)\u00b1\u2212342\u00d73=2\u00b134i23<\/h3>\n

9. Solve the equation
\nx2+x+12\u2013\u221a=0x2+x+12=0
\nAnd evaluate<\/h2>\n

Ans:
\nQuadratic equation x2+x+12\u2013\u221a=0x2+x+12=0
\nGeneral form ax2+bx+c=0ax2+bx+c=0
\nWe obtain a=2\u2013\u221a,b=2\u2013\u221a,andc=1a=2,b=2,andc=1
\nTherefore, the discriminant of the given equation is
\nD=b2\u22124ac=(2\u2013\u221a)2\u22124\u00d72\u2013\u221a\u00d71=2-42\u2013\u221aD=b2\u22124ac=(2)2\u22124\u00d72\u00d71=2-42
\nTherefore, the required solutions are
\n=\u2212b\u00b1D\u2212\u2212\u221a2a=\u2212(2\u2013\u221a)\u00b12\u221242\u2013\u221a\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u221a2\u00d72\u2013\u221a=\u22121\u00b1(22\u2013\u221a\u22121\u2212\u2212\u2212\u2212\u2212\u2212\u221a)i2=\u2212b\u00b1D2a=\u2212(2)\u00b12\u2212422\u00d72=\u22121\u00b1(22\u22121)i2<\/h3>\n

10. Solve the equation
\nx2+x2\u2013\u221a+1=0x2+x2+1=0
\nAnd evaluate<\/h2>\n

Ans:
\nQuadratic equation x2+x2\u2013\u221a+1=0x2+x2+1=0
\nGeneral form ax2+bx+c=0ax2+bx+c=0
\nWe obtain a=2\u2013\u221a,b=1,andc=2\u2013\u221aa=2,b=1,andc=2
\nTherefore, the discriminant of the given equation is
\nD=b2\u22124ac=(1)2\u22124\u00d72\u2013\u221a\u00d72\u2013\u221a=\u22127D=b2\u22124ac=(1)2\u22124\u00d72\u00d72=\u22127
\nTherefore, the required solutions are
\n=\u2212b\u00b1D\u2212\u2212\u221a2a=\u2212(1)\u00b1\u22127\u2212\u2212\u2212\u221a2\u00d72\u2013\u221a=\u22121\u00b17\u2013\u221ai22\u2013\u221a=\u2212b\u00b1D2a=\u2212(1)\u00b1\u221272\u00d72=\u22121\u00b17i22<\/h3>\n","protected":false},"excerpt":{"rendered":"

Exercise 5.1 1. Express the given complex number in the form a+ib:(5i)(\u221235i)a+ib:(5i)(\u221235i) And evaluate Ans: Evaluate the complex number (5i)(\u221235i)=\u22125\u00d735\u00d7i\u00d7i(5i)(\u221235i)=\u22125\u00d735\u00d7i\u00d7i (5i)(\u221235i)=\u22123i2\u22ef[i2=\u22121](5i)(\u221235i)=\u22123i2\u22ef[i2=\u22121] (5i)(\u221235i)=3(5i)(\u221235i)=3 2. Express the given complex number in the form a+ib:i9+i19a+ib:i9+i19 And evaluate Ans: Evaluate the complex number i9+i19=i4\u00d72+1+i4\u00d74+3i9+i19=i4\u00d72+1+i4\u00d74+3 i9+i19=(i4)2.i+(i4)4.i3\u22ef[i4=1,i3=\u22121]i9+i19=(i4)2.i+(i4)4.i3\u22ef[i4=1,i3=\u22121] i9+i19=0i9+i19=0 3. Express the given complex number in the form a+ib:i\u221239a+ib:i\u221239 And evaluate Ans: […]<\/p>\n","protected":false},"author":21830,"featured_media":121167,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[7],"tags":[],"class_list":{"0":"post-121815","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-education"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/121815","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/users\/21830"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/comments?post=121815"}],"version-history":[{"count":1,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/121815\/revisions"}],"predecessor-version":[{"id":121816,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/121815\/revisions\/121816"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media\/121167"}],"wp:attachment":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media?parent=121815"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/categories?post=121815"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/tags?post=121815"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}