RD Chapter 14 Quadratic Equations Ex 14.1 Solutions
Question - 1 : - Solve the following quadratic equations by factorization method only:
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