Question

Which of the following is a quadratic equation?

• A. $x^2+2x+1={(4-x)}^2+3$
• B. $-2x^2=(5-x)(2x-\frac{2}{5})$
• C. $(k+1)x^2+\frac{3}{2}x=7$ ( Where k =-1 )
• D. $x^3-x^2={(x-1)}^3$

Answer 1

The correct option is D $x^3-x^2={(x-1)}^3$

The correct answer is option $\left(D\right)$

Main concept used : An equation of the form $a{x}^2+bx+c=0$ where, $a,b,c,$ are numbers and $a\ne 0$, is called a quadratic equation.

$\left(a\right)$ $x^2+2x+1={(4-x)}^2+3$

$\Rightarrow x^2+2x+1=(4{)}^2+(x{)}^2-2\left(4\right)\left(x\right)+3$

$\Rightarrow 2x+1=16-8x+3$

$\therefore$ Coefficient of $x^2$ is zero or a = 0. So, it is not a quadratic equation.

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

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

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

$\Rightarrow 0=10x-2+\frac{2}{5}x$

As the coefficient of $x^2$ in the above equation is zero or $a=0.$

So, it is not a quadratic equation.

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

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

So, the coefficient of $x^2$ is zero or $a=0$. Hence, the equation in not quadratic.

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

$\Rightarrow x^3-x^2=(x{)}^3-(1{)}^3-3\left(x{)}^2\right(1)+3(x\left)\right(1{)}^2$

$\Rightarrow x^3-x^2=x^3-1-3x^2+3x$

$\Rightarrow -x^2=-1-3x^2+3x$

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

As the coefficient of $x^2$ in the above equation is $2$ or $a=2$, so it is a quadratic equation.