Question

Check whether the following are quadratic equations :
(i) $(x + 1)^2 = 2(x – 3)$ (ii) $x^2 – 2x = (–2) (3 – x)$
(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) $x^2 + 3x + 1 = (x – 2)^2$
(vii) $(x + 2)^3 = 2x (x^2 – 1)$ (viii) $x^3 – 4x^2 – x + 1 = (x – 2)^3$

Answer 1

(i) $(x + 1)^2 = 2(x – 3) ⇒ x^2 + 2x + 1 = 2x -6 ⇒ x^2 +7 =0$
It is of the form $ax^2 + bx +c =0.$

Hence, the given equation is a quadratic equation.

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**(ii) **$x^2 – 2x = (–2) (3 – x) ⇒ x^2 -2x = -6 + 2x ⇒ x^2 -4x +6$
It is of the form $ax^2 + bx +c =0.$

Hence, the given equation is a quadratic equation.

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(iii) $(x – 2)(x + 1) = (x – 1)(x + 3) ⇒ x^2 -x-2 = x^2 +2x -3 ⇒ 3x-1$
It is not of the form $ax^2 + bx +c =0.$

Hence, the given equation is not a quadratic equation.

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(iv) $(x – 3)(2x +1) = x(x + 5) ⇒ 2x^2 -5x -3 = x^2 +5x ⇒ x^2 -10x -3$
It is of the form $ax^2 + bx +c =0.$

Hence, the given equation is a quadratic equation.

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(v) $(2x – 1)(x – 3) = (x + 5)(x – 1) ⇒ 2x^2 -7x +3 = x^2 +4x -5 ⇒ x^2-11x +8 =0$
It is of the form $ax^2 + bx +c =0.$

Hence, the given equation is a quadratic equation.

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(vi) $x^2 + 3x + 1 = (x – 2)^2 ⇒ x^2 +3x +1 = x^2 +4 -4x ⇒7x -3 =0$
It is not of the form $ax^2 + bx +c =0.$

Hence, the given equation is not a quadratic equation.

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(vii) $(x + 2)^3 = 2x (x^2 – 1) ⇒x^3 +8 +6x^2 +12x ⇒ 2x^3 -2x ⇒ x^2 -14x -6x^2 -8=0$
It is of the form $ax^2 + bx +c =0.$

Hence, the given equation is a quadratic equation.

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**(viii) **$x^3 – 4x^2 – x + 1 = (x – 2)^3 ⇒ x^3 -4x^2 -x +1 = x^3 -8-6x^2 +12x ⇒ 2x^2 -13x +9$
It is of the form $ax^2 + bx +c =0.$

Hence, the given equation is a quadratic equation.