# Which of the following is a quadratic equation?

Question:

Which of the following is a quadratic equation?

(a) x2 + 2x + 1 = (4 – x)2 + 3

(b) $-2 x^{2}=(5-x)\left(2 x-\frac{2}{5}\right.$

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

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

Solution:

(a) Given that, $\quad x^{2}+2 x+1=(4-x)^{2}+3$

$\Rightarrow \quad x^{2}+2 x+1=16+x^{2}-8 x+3$

$\Rightarrow \quad 10 x-18=0$

which is not of the form $a x^{2}+b x+c, a \neq 0$. Thus, the equation is not a quadratic equation.

(b) Given that, $-2 x^{2}=(5-x)\left(2 x-\frac{2}{5}\right)$

$\Rightarrow$ $-2 x^{2}=10 x-2 x^{2}-2+\frac{2 x}{5}$

$\Rightarrow \quad 50 x+2 x-10=0$

$\Rightarrow \quad 52 x-10=0$

which is also not a quadratic equation.

(c) Given that, $x^{2}(k+1)+\frac{3}{2} x=7$

Given, $k=-1$

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

$\Rightarrow \quad 3 x-14=0$

which is also not a quadratic equation.

(d) Given that, $x^{3}-x^{2}=(x-1)^{3}$

$\rightarrow \quad x^{3}-x^{2}=x^{3}-3 x^{2}(1)+3 x(1)^{2}-(1)^{3}$

$\left[\because(a-b)^{3}=a^{3}-b^{3}+3 a b^{2}-3 a^{2} b\right]$

$\Rightarrow \quad x^{3}-x^{2}=x^{3}-3 x^{2}+3 x-1$

$\Rightarrow \quad-x^{2}+3 x^{2}-3 x+1=0$

$\Rightarrow \quad 2 x^{2}-3 x+1=0$

which represents a quadratic equation because it has the quadratic rom $a x^{2}+b x+c=0, a \neq 0$