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RD Chapter 14- Quadratic Equations Ex-14.1 Interview Questions Answers

Question 1 : Solve the following quadratic equations by factorization method only:

x2 + 1 = 0

Answer 1 :

Given: x2 +1 = 0

We know, i2 =–1  1 = –i2

By substituting 1 = –i2 inthe above equation, we get

x2 – i2 =0

[Byusing the formula, a2 – b2 = (a + b) (a –b)]

(x + i) (x – i) = 0

x + i = 0 or x – i = 0

x = –i or x = i

The roots of thegiven equation are i, -i

Question 2 :

9x2 + 4 = 0

Answer 2 :

Given: 9x2 +4 = 0

9x2 +4 × 1 = 0

We know, i2 =–1  1 = –i2

By substituting 1 = –i2 inthe above equation, we get

So,

9x2 +4(–i2) = 0

9x2 –4i2 = 0

(3x)2 –(2i)2 = 0

[Byusing the formula, a2 – b2 = (a + b) (a –b)]

(3x + 2i) (3x – 2i) =0

3x + 2i = 0 or 3x – 2i= 0

3x = –2i or 3x = 2i

x = -2i/3 or x = 2i/3

The roots of thegiven equation are 2i/3, -2i/3

Question 3 :

x2 + 2x + 5 = 0

Answer 3 :

Given: x2 +2x + 5 = 0

x2 +2x + 1 + 4 = 0

x2 +2(x) (1) + 12 + 4 = 0

(x + 1)2 +4 = 0 [since, (a + b)2 = a2 + 2ab + b2]

(x + 1)2 +4 × 1 = 0

We know, i2 =–1  1 = –i2

By substituting 1 = –i2 inthe above equation, we get

(x + 1)2 +4(–i2) = 0

(x + 1)2 –4i2 = 0

(x + 1)2 –(2i)2 = 0

[Byusing the formula, a2 – b2 = (a + b) (a –b)]

(x + 1 + 2i)(x + 1 –2i) = 0

x + 1 + 2i = 0 or x +1 – 2i = 0

x = –1 – 2i or x = –1+ 2i

The roots of thegiven equation are -1+2i, -1-2i

Question 4 :

4x2 – 12x + 25 = 0

Answer 4 :

Given: 4x2 –12x + 25 = 0

4x2 –12x + 9 + 16 = 0

(2x)2 –2(2x)(3) + 32 + 16 = 0

(2x – 3)2 +16 = 0 [Since, (a + b)2 = a2 + 2ab + b2]

(2x – 3)2 +16 × 1 = 0

We know, i2 =–1  1 = –i2

By substituting 1 = –i2 inthe above equation, we get

(2x – 3)2 +16(–i2) = 0

(2x – 3)2 –16i2 = 0

(2x – 3)2 –(4i)2 = 0

[Byusing the formula, a2 – b2 = (a + b) (a – b)]

(2x – 3 + 4i) (2x – 3– 4i) = 0

2x – 3 + 4i = 0 or 2x– 3 – 4i = 0

2x = 3 – 4i or 2x = 3+ 4i

x = 3/2 – 2i or x =3/2 + 2i

The roots of thegiven equation are 3/2 + 2i, 3/2 – 2i

Question 5 :

x2 + x + 1 = 0

Answer 5 :

Given: x2 +x + 1 = 0

x2 + x+ ¼ + ¾ = 0

x2 + 2(x) (1/2) + (1/2)2 + ¾ = 0

(x + 1/2)2 +¾ = 0 [Since, (a + b)2 = a2 + 2ab + b2]

(x + 1/2)2 +¾ × 1 = 0

We know, i2 =–1  1 = –i2

By substituting 1 = –i2 inthe above equation, we get

(x + ½)2 +¾ (-1)2 = 0

(x + ½)2 +¾ i2 = 0

(x + ½)2 –(√3i/2)2 = 0

[Byusing the formula, a2 – b2 = (a + b) (a – b)]

(x + ½ + √3i/2) (x + ½ – √3i/2) = 0

(x + ½ + √3i/2) = 0 or (x + ½ – √3i/2) = 0

x = -1/2 – √3i/2 or x = -1/2 + √3i/2

The roots of thegiven equation are -1/2 + √3i/2,-1/2 – √3i/2

Question 6 :

4x2 + 1 = 0

Answer 6 :

Given: 4x2 +1 = 0

We know, i2 =–1  1 = –i2

By substituting 1 = –i2 inthe above equation, we get

4x2 –i2 = 0

(2x)2 –i2 = 0

[Byusing the formula, a2 – b2 = (a + b) (a – b)]

(2x + i) (2x – i) = 0

2x + i = 0 or 2x – i =0

2x = –i or 2x = i

x = -i/2 or x = i/2

The roots of thegiven equation are i/2, -i/2

Question 7 :

x2 – 4x + 7 = 0

Answer 7 :

Given: x2 –4x + 7 = 0

x2 –4x + 4 + 3 = 0

x2 –2(x) (2) + 22 + 3 = 0

(x – 2)2 +3 = 0 [Since, (a – b)2 = a2 – 2ab + b2]

(x – 2)2 +3 × 1 = 0

We know, i2 =–1  1 = –i2

By substituting 1 = –i2 inthe above equation, we get

(x – 2)2 +3(–i2) = 0

(x – 2)2 –3i2 = 0

(x – 2)2 –(√3i)2 = 0

[Byusing the formula, a2 – b2 = (a + b) (a – b)]

(x – 2 + √3i) (x – 2 – √3i) = 0

(x – 2 + √3i) = 0 or (x – 2 – √3i) = 0

x = 2 – √3i or x = 2 + √3i

x = 2 ± √3i

The roots of thegiven equation are 2 ± √3i

Question 8 :

x2 + 2x + 2 = 0

Answer 8 :

Given: x2 +2x + 2 = 0

x2 +2x + 1 + 1 = 0

x2 +2(x)(1) + 12 + 1 = 0

(x + 1)2 +1 = 0 [ (a + b)2 =a2 + 2ab + b2]

We know, i2 =–1  1 = –i2

By substituting 1 = –i2 inthe above equation, we get

(x + 1)2 +(–i2) = 0

(x + 1)2 –i2 = 0

(x + 1)2 –(i)2 = 0

[Byusing the formula, a2 – b2 = (a + b) (a – b)]

(x + 1 + i) (x + 1 –i) = 0

x + 1 + i = 0 or x + 1– i = 0

x = –1 – i or x = –1 +i

x = -1 ± i

The roots of thegiven equation are -1 ± i

Question 9 :

5x2 – 6x + 2 = 0

Answer 9 :

Given: 5x2 –6x + 2 = 0

We shall applydiscriminant rule,

Where, x = (-b ±√(b2 – 4ac))/2a

Here, a = 5, b = -6, c= 2

So,

x = (-(-6) ±√(-62 – 4(5)(2)))/ 2(5)

= (6 ± √(36-40))/10

= (6 ± √(-4))/10

= (6 ± √4(-1))/10

We have i2 =–1

By substituting –1 = i2 inthe above equation, we get

x = (6 ± √4i2)/10

= (6 ± 2i)/10

= 2(3±i)/10

= (3±i)/5

x = 3/5 ± i/5

The roots of thegiven equation are 3/5 ± i/5

Question 10 :

21x2 + 9x + 1 = 0

Answer 10 :

Given: 21x2 +9x + 1 = 0

We shall applydiscriminant rule,

Where, x = (-b ±√(b2 – 4ac))/2a

Here, a = 21, b = 9, c= 1

So,

x = (-9 ±√(92 – 4(21)(1)))/ 2(21)

= (-9 ± √(81-84))/42

= (-9 ± √(-3))/42

= (-9 ± √3(-1))/42

We have i2 =–1

By substituting –1 = i2 inthe above equation, we get

x = (-9 ± √3i2)/42

= (-9 ± √(√3i)2/42

= (-9 ± √3i)/42

= -9/42 ± √3i/42

= -3/14 ± √3i/42

The roots of thegiven equation are -3/14 ± √3i/42


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RD Chapter 14- Quadratic Equations Ex-14.1 Contributors

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