(i) Given,
(x + 1)2 = 2(x – 3)
By using the formula for (a+b)2 = a2+2ab+b2
⇒ x2 + 2x + 1 = 2x – 6
⇒ x2 + 7 = 0
Since the above equation is in the form of ax2 + bx + c = 0.
Therefore, the given equation is quadratic equation.
(ii) Given, x2 – 2x = (–2) (3 – x)
By using the formula for (a+b)2 = a2+2ab+b2
⇒ x2 – 2x = -6 + 2x
⇒ x2 – 4x + 6 = 0
Since the above equation is in the form of ax2 + 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 = a2+2ab+b2
⇒ x2 – x – 2 = x2 + 2x – 3
⇒ 3x – 1 = 0
Since the above equation is not in the form of ax2 + 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=a2+2ab+b2
⇒ 2x2 – 5x – 3 = x2 + 5x
⇒ x2 – 10x – 3 = 0
Since the above equation is in the form of ax2 + 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=a2+2ab+b2
⇒ 2x2 – 7x + 3 = x2 + 4x – 5
⇒ x2 – 11x + 8 = 0
Since the above equation is in the form of ax2 + bx + c = 0.
Therefore, the given equation is quadratic equation.
(vi) Given, x2 + 3x + 1 = (x – 2)2
By using the formula for (a+b)2=a2+2ab+b2
⇒ x2 + 3x + 1 = x2 + 4 – 4x
⇒ 7x – 3 = 0
Since the above equation is not in the form of ax2 + bx + c = 0.
Therefore, the given equation is not a quadratic equation.
(vii) Given, (x + 2)3 = 2x(x2 – 1)
By using the formula for (a+b)2 = a2+2ab+b2
⇒ x3 + 8 + x2 + 12x = 2x3 – 2x
⇒ x3 + 14x – 6x2 – 8 = 0
Since the above equation is not in the form of ax2 + bx + c = 0.
Therefore, the given equation is not a quadratic equation.
(viii) Given, x3 – 4x2 – x + 1 = (x – 2)3
By using the formula for (a+b)2 = a2+2ab+b2
⇒ x3 – 4x2 – x + 1 = x3 – 8 – 6x2 + 12x
⇒ 2x2 – 13x + 9 = 0
Since the above equation is in the form of ax2 + bx + c = 0.
Therefore, the given equation is quadratic equation.
