RD Chapter 14 Quadratic Equations Ex 14.1 Solutions
Question - 11 : - x2 – x + 1 = 0
Answer - 11 : -
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 (-i)2 = 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 - 12 : - 17x2 – 8x + 1 = 0
Answer - 12 : -
Given: 17x2 –8x + 1 = 0
We shall applydiscriminant rule,
Where, x = (-b ±√(b2 – 4ac))/2a
Here, a = 17, b = -8,c = 1
So,
x = (-(-8) ±√(-82 – 4(17)(1)))/ 2(17)
= (8 ± √(64-68))/34
= (8 ± √(-4))/34
= (8 ± √4(-1))/34
We have i2 =–1
By substituting –1 = i2 inthe above equation, we get
x = (8 ± √(2i)2)/34
= (8 ± 2i)/34
= 2(4±i)/34
= (4±i)/17
x = 4/17 ± i/17
∴ The roots of thegiven equation are 4/17 ± i/17