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Class 11 NCERT Solutions- Chapter 5 Complex Numbers And Quadratic Equations – Exercise 5.3
  • Last Updated : 10 Mar, 2021

Solve each of the following equations:

Question 1. x2 + 3 = 0

Solution:

We have,

x2 + 3 = 0 or x2 + 0 × x + 3 = 0 —(1)

Discriminant, D = b2 – 4ac

from (1) , a = 1 , b = 0 and c = 3



D = (0)2 – 4*(1)*(3)

D = -12 

Since , x = \frac{( -b  ± \sqrt{D} }{2a}

Therefore , 

x = \frac{( -0  ± \sqrt{-12} }{2(1)}

x = \frac{(± 2\sqrt{3}i }{2}

x = ± √3 i

Hence , the solution of x2 + 3 = 0 is ± √3 i



Question 2. 2x2 + x + 1 = 0

Solution:

We have,

2x2 + x + 1 = 0 —(1)

Discriminant, D = b2 – 4ac

from (1) , a = 2 , b = 1 and c = 1

D = (1)2 – 4*(2)*(1)

D = -7

Since , x = \frac{( -b  ± \sqrt{D} }{2a}

Therefore ,

x = \frac{( -1  ± \sqrt{7} i}{2(2)}

x = \frac{( -1  ± \sqrt{7} i}{4}

Hence , the solution of  2x2 + x + 1 = 0  is \frac{( -1  ± \sqrt{7} i}{4}  .

Question 3. x2 + 3x + 9 = 0

Solution:

We have,

x2 + 3x + 9 = 0 —(1)

Discriminant, D = b2 – 4ac

from (1) , a = 1 , b = 3 and c = 9

D = (3)2 – 4*(1)*(9)

D = -27

Since , x = \frac{( -b  ± \sqrt{D} }{2a}

Therefore ,

x = \frac{( -3  ± \sqrt{27}i }{2(1)}

x = \frac{( -3  ± 3\sqrt{3}i }{2}

Hence , the solution of  x2 + 3x + 1 = 0  is \frac{( -3  ± 3\sqrt{3}i }{2}  .

Question 4. -x2+ x – 2 = 0

Solution:

We have,

-x2 + x – 2 = 0 —(1)

Discriminant, D = b2 – 4ac

from (1) , a = -1 , b = 1 and c = -2

D = (1)2 – 4*(-1)*(-2)

D = -7

Since , x = \frac{ -b  ± \sqrt{D} }{2a}

Therefore ,

x = \frac{ -1  ± \sqrt{7} i}{2(-1)}

x = \frac{ -1  ± \sqrt{7} i}{-2}

Hence , the solution of  -x2 + x – 2= 0  is \frac{ -1  ± \sqrt{7} i}{-2}  .

Question 5. x2 + 3x + 5 = 0

Solution:

We have,

x2 + 3x + 5 = 0 —(1)

Discriminant, D = b2 – 4ac

from (1) , a = 1 , b = 3 and c = 5

D = (3)2 – 4*(1)*(5)

D = -11

Since , x = \frac{ -b  ± \sqrt{D} }{2a}

Therefore ,

x = \frac{ 3  ± \sqrt{11}i }{2(1)}

x = \frac{ 3  ± \sqrt{11}i }{2}

Hence , the solution of  -x2 + x – 2= 0  is \frac{ 3  ± \sqrt{11}i }{2}  .

Question 6. x2 – x + 2 = 0

Solution:

We have,

x2 – x + 2 = 0 —(1)

Discriminant, D = b2 – 4ac

from (1) , a = 1 , b = -1 and c = 2

D = (-1)2 – 4*(1)*(2)

D =  -7

Since , x = \frac{ -b  ± \sqrt{D} }{2a}

Therefore ,

x = \frac{ -(-1)  ± \sqrt{-7} }{2(1)}

x = \frac{ 1  ± \sqrt{7} }{2}

Hence , the solution of  -x2 + x – 2= 0  is \frac{ 1  ± \sqrt{7} }{2}  .

Question 7. √2x2 + x + √2 = 0

Solution:

We have,

√2x2 + x + √2 = 0 —(1)

Discriminant, D = b2 – 4ac

from (1) , a = √2 , b = 1 and c = √2

D = (1)2 – 4*(√2)*(√2)

D =  -7

Since , x = \frac{ -b  ± \sqrt{D} }{2a}

Therefore ,

x = \frac{ -1 ± \sqrt{7} i}{ 2(\sqrt{2})}

Hence , the solution of  -x2 + x – 2= 0  is \frac{ -1 ± \sqrt{7} i}{ 2\sqrt{2}}  .

Question 8. √3x2  – √2x + 3√3 = 0

Solution:

We have,

√3x2 – √2x + 3√3 = 0 —(1)

Discriminant, D = b2 – 4ac

from (1) , a = √3 , b = -√2 and c = 3√3

D = (-√2)2 – 4*(√3)*(3√3)

D =  -34

Since , x = \frac{ -b  ± \sqrt{D} }{2a}

Therefore ,

x = \frac{-(-\sqrt{2} ± \sqrt{34}i)}{ 2(\sqrt{3})}

x =  \frac{\sqrt{2} ± \sqrt{34}i}{ 2(\sqrt{3})}

Hence , the solution of  -x2 + x – 2= 0  is \frac{\sqrt{2} ± \sqrt{34}i}{ 2(\sqrt{3})} .

Question 9. x2 + x + \frac{1}{\sqrt{2}}   = 0

Solution:

We have,

x2 + x + \frac{1}{\sqrt{2}}  = 0  or  √2x2 + √2x + 1 = 0 —(1)

Discriminant, D = b2 – 4ac

from (1) , a = √2 , b = √2 and c = 1

D = (√2)2 – 4*(√2)*(1)

D =  2 – 4√2 = 2 ( 1 – 2√2 )

Since , x = \frac{ -b  ± \sqrt{D} }{2a}

Therefore ,

x = \frac{ -\sqrt{2}  ± \sqrt{2(1-2\sqrt{2}})}{2\sqrt{2}}

x = \frac{ -1  ± \sqrt{2(\sqrt{2}-1)}i}{2}

x = \frac{ -1  ± \sqrt{(2\sqrt{2}-1}i}{2}

Hence , the solution of  -x2 + x – 2= 0  is \frac{ -1  ± \sqrt{(2\sqrt{2}-1}i}{2} .

Question 10. x2 \frac{x }{ \sqrt{2} }  + 1 = 0

Solution:

We have,

x2\frac{x }{ \sqrt{2} }  + 1 = 0  or √2x2 + x + √2 = 0 —(1)

Discriminant, D = b2 – 4ac

from (1) , a = √2 , b = 1 and c = √2

D = (1)2 – 4*(√2)*(√2)

D =  -7

Since , x = \frac{ -b  ± \sqrt{D} }{2a}

Therefore ,

x = \frac{ -1  ± \sqrt{7}i }{2 (\sqrt{2})}

x = \frac{ -1  ± \sqrt{7}i }{2\sqrt{2}}

Hence , the solution of  x2\frac{x }{ \sqrt{2} }  + 1 = 0  is \frac{ -1  ± \sqrt{7}i }{2\sqrt{2}} .

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