How i solve the quadratic equation of exercise 4.1 is easy method. It is very important type of question for board exam. This is my favorite question of math 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 – 4×2 – x + 1 = (x – 2)3

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# 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 – 4×2 – x + 1 = (x – 2)3 Q.1

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(i) Given,

(x + 1)

^{2}= 2(x – 3)By using the formula for (a+b)

^{2 }= a^{2}+2ab+b^{2}⇒ x

^{2}+ 2x + 1 = 2x – 6⇒ x

^{2}+ 7 = 0Since the above equation is in the form of ax

^{2}+ bx + c = 0.Therefore, the given equation is quadratic equation.

(ii) Given, x

^{2}– 2x = (–2) (3 – x)By using the formula for (a+b)

^{2 }= a^{2}+2ab+b^{2}⇒ x

^{2 }–^{ }2x = -6 + 2x⇒ x

^{2 }– 4x + 6 = 0Since the above equation is in the form of ax

^{2}+ bx + c = 0.Therefore, the given equation is quadratic equation.

(iii) Given, (x – 2)(x + 1) = (x – 1)(x + 3)

By using the formula for (a+b)

^{2 }= a^{2}+2ab+b^{2}⇒ x

^{2 }– x – 2 = x^{2 }+ 2x – 3⇒ 3x – 1 = 0

Since the above equation is not in the form of ax

^{2}+ bx + c = 0.Therefore, the given equation is not a quadratic equation.

(iv) Given, (x – 3)(2x +1) = x(x + 5)

By using the formula for (a+b)

^{2}=a^{2}+2ab+b^{2}⇒ 2x

^{2 }– 5x – 3 = x^{2 }+ 5x⇒ x

^{2 }– 10x – 3 = 0Since the above equation is in the form of ax

^{2}+ bx + c = 0.Therefore, the given equation is quadratic equation.

(v) Given, (2x – 1)(x – 3) = (x + 5)(x – 1)

By using the formula for (a+b)

^{2}=a^{2}+2ab+b^{2}⇒ 2x

^{2 }– 7x + 3 = x^{2 }+ 4x – 5⇒ x

^{2 }– 11x + 8 = 0Since the above equation is in the form of ax

^{2}+ bx + c = 0.Therefore, the given equation is quadratic equation.

(vi) Given, x

^{2}+ 3x + 1 = (x – 2)^{2}By using the formula for (a+b)

^{2}=a^{2}+2ab+b^{2}⇒ x

^{2}+ 3x + 1 = x^{2}+ 4 – 4x⇒ 7x – 3 = 0

Since the above equation is not in the form of ax

^{2}+ bx + c = 0.Therefore, the given equation is not a quadratic equation.

(vii) Given, (x + 2)

^{3}= 2x(x^{2}– 1)By using the formula for (a+b)

^{2 }= a^{2}+2ab+b^{2}⇒ x

^{3}+ 8 + x^{2}+ 12x = 2x^{3}– 2x⇒ x

^{3}+ 14x – 6x^{2}– 8 = 0Since the above equation is not in the form of ax

^{2}+ bx + c = 0.Therefore, the given equation is not a quadratic equation.

(viii) Given, x

^{3}– 4x^{2}– x + 1 = (x – 2)^{3}By using the formula for (a+b)

^{2 }= a^{2}+2ab+b^{2}⇒ x

^{3}– 4x^{2}– x + 1 = x^{3}– 8 – 6x^{2 }+ 12x⇒ 2x

^{2}– 13x + 9 = 0Since the above equation is in the form of ax

^{2}+ bx + c = 0.Therefore, the given equation is quadratic equation.