Q:-12 The value of a, if (x - a) is a factor of (x² - ax2 + 2x + 9 - 1
Answers
Answer:
Check whether the following are quadratic equations:
(i) (x + 1)2 = 2(x – 3)
(ii) x2 – 2x = (–2)(3 – x)
(iii) (x – 2)(x + 1) = (x – 1)(x + 3)
(iv) (x – 3)(2x + 1) = x(x + 5)
(v) (2x – 1) (x – 3) – (x + 5) (x – 1)
(vi) x2 + 3x +1 = (x – 2)2
(vii) (x + 2)3 = 2x(x2 – 1)
(viii) x3 – 4x2 – × + 1 = (x – 2)3
Sol. (i) (x + 1)2 = 2(x – 3)
We have:
(x + 1)2 = 2 (x – 3) x2 + 2x + 1 = 2x – 6
⇒ x2 + 2x + 1 – 2x + 6 = 0
⇒ x2 + 70
Since x2 + 7 is a quadratic polynomial
∴ (x + 1)2 = 2(x – 3) is a quadratic equation.
(ii) x2– 2x = (–2) (3 – x)
We have:
x2 – 2x = (– 2) (3 – x)
⇒ x2 – 2x = –6 + 2x
⇒ x2 – 2x – 2x + 6 = 0
⇒ x2 – 4x + 6 = 0
Since x2 – 4x + 6 is a quadratic polynomial
∴ x2 – 2x = (–2) (3 – x) is a quadratic equation.
(iii) (x – 2) (x + 1) = (x – 1) (x + 3)
We have:
(x – 2) (x + 1) = (x – 1) (x + 3)
⇒ x2 – x – 2 = x2 + 2x – 3
⇒ x2 – x – 2 – x2 – 2x + 3 = 0
⇒ –3x + 1 = 0
Since –3x + 1 is a linear polynomial
∴ (x – 2) (x + 1) = (x – 1) (x + 3) is not quadratic equation.
(iv) (x – 3) (2x + 1) = x(x + 5)
We have:
(x – 3) (2x + 1) = x(x + 5)
⇒ 2x2 + x – 6x – 3 = x2 + 5x
⇒ 2x2 – 5x – 3 – x2 – 5x – 0
⇒ x2 + 10x – 3 = 0
Since x2 + 10x – 3 is a quadratic polynomial
∴ (x – 3) (2x + 1) = x(x + 5) is a quadratic equation.
(v) (2x – 1) (x – 3) = (x + 5) (x – 1)
We have:
(2x – 1) (x – 3) = (x + 5) (x – 1)
⇒ 2x2 – 6x – x + 3 = x2 – x + 5x – 5
⇒ 2x2 – x2 – 6x – x + x – 5x + 3 + 5 = 0
⇒ x2 – 11x + 8 = 0
Since x2 – 11x + 8 is a quadratic polynomial
∴ (2x – 1) (x – 3) = (x + 5) (x – 1) is a quadratic equation.
(vi) x2 + 3x + 1 = (x – 2)2
We have:
x2 + 3x + 1 = (x – 2)2
⇒ x2 + 3x + 1 = x2 – 4x + 4
⇒ x2 + 3x + 1 – x2 + 4x – 4 =0
⇒ 7x – 3 = 0
Since 7x – 3 is a linear polynomial.
∴ x2 + 3x + 1 = (x – 2)2 is not a quadratic equation.
(vii) (x + 2)3 = 2x(x2 – 1)
We have:
(x + 2)3 = 2x(x2 – 1)
x3 + 3x2(2) + 3x(2)2 + (2)3 = 2x3 – 2x
⇒ x3 + 6x2 + 12x + 8 = 2x3 – 2x
⇒ x3 + 6x2 + 12x + 8 – 2x3 + 2x = 0
⇒ –x3 + 6x2 + 14x + 8 = 0
Since –x3 + 6x2 + 14x + 8 is a polynomial of degree 3
∴ (x + 2)3 = 2x(x2 – 1) is not a quadratic equation.
(viii) x3 – 4x2 – x + 1 = (x – 2)3
We have:
x3 – 4x2 – x + 1 = (x – 2)3
⇒ x3 – 4x2 – x + 1 = x3 + 3x2(– 2) + 3x(– 2)2 + (– 2)3
⇒ x3 – 4x2 – x + 1 = x3 – 6x2 + 12x – 8
⇒ x3 – 4x2 – x – 1 – x3 + 6x2 – 12x + 8 = 0
2x2 – 13x + 9 = 0
Since 2x2 – 13x + 9 is a quadratic polynomial
∴ x3 – 4x2 – x + 1 = (x – 2)3 is a