#### Choose the correct answer from the given four options in the following questions: Which of the following is a quadratic equation?$\\(A) x^{2}+2x+1=(4-x)^{2}+3\\ (B)-2x^{2}=(5-x)\left ( 2x-\frac{2}{5} \right )\\ (C)(k+1)x^{2}+\frac{3}{2}x=7\; \; where\;\; k=-1\\ (D)x^{3}-x^{2}=(x-1)^{3}$

Answer(D)$x^{3} - x^{2} = (x - 1)^{3}$

Solution:

A quadratic equation in x is an equation that can be written in the standard form.

$ax^{2} + bx + c = 0$ where a, b and c are real number with $a \neq 0$
$\\(A) x^{2} + 2x + 1 = (4 - x)^{2} + 3\\ x^{2} + 2x + 1 = (4)^{2} + (x)^{2} - 2(4)(x) + 3\;\;\;\; (Using (a - b)^{2} = a^{2} + b^{2} - 2ab)\\ x^{2} + 2x + 1 = 16 + x^{2} - 8x + 3\\ x^{2} + 2x + 1 - 16 - x^{2} + 8x - 3 = 0\\ 10x - 18 = 0$

$10x - 18 = 0$  is not in the form of $ax^{2} + bx + c = 0$

Hence it is not a quadratic equation.

(B)

$\\-2x^{2}=(5-x)\left ( 2x-\frac{2}{5} \right )\\ -2x^{2}=10x-2-2x^{2}+\frac{2}{5}x\\ 10x+\frac{2}{5}x-2=0\\ \frac{50x+2x-10}{5}=0\\ 5x-10=0$

This is not in the form of $ax^{2} + bx + c = 0$

Hence it is not a quadratic equation

$(C)(k+1)x^{2}+\frac{3}{2}x=7\; \; where\;\; k=-1\\$

Put       k = – 1

$\\((-1)+1)x^{2}+\frac{3}{2}x=7\\ \frac{3}{2}x-7=0\\ \frac{3x-14}{2}=0\\ 3x-14=0$

This is not in the form of $ax^{2} + bx + c = 0$

Hence this is not a quadratic equation.

(D)

$\\x^{3}-x^{2}=(x - 1)^{3}\\ x^{3}-x^{2} =(x)^{3} - (1)^{3} (1) + 3(x)(1)^{2}\;\;\;\; [using\; (a - b)^{3} = a^{3} - b^{3} - 3a^{2}b + 3ab^{2}]\\ x^{3} - x^{2} - x^{3} + 1 + 3x^{2} - 3x = 0\\ 2x^{2} - 3x + 1 = 0$

This is in the form of $ax^{2} + bx + c = 0$

Hence $x^{3}-x^{2}=(x-1)^{3}$ is a quadratic equation.