Question

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Answer 1

Answer:

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Answer 2

Answer:

(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)