Question

Question 1
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) $(k+1)x^2+\frac{3}{2}x=7,wherek=-1$
(d) $x^3-x^2=(x-1{)}^3$

Answer 1

The answer is (d).
a) Given that, $x^2+2x+1=(4-x{)}^2+3$
$\Rightarrow x^2+2x+1=16+x^2-8x+3$
$\Rightarrow 10x-18=0$
Which is not of the form $a{x}^2+bx+c$, a $\ne$ 0. Thus the equation is not a quadratic equation.
b) Given that, $-2x^2=(5-x)(2x-\frac{2}{5})$
$\Rightarrow -2x^2=10x-2x^2-2+\frac{2x}{5}$
$\Rightarrow 50x+2x-10=0$
$\Rightarrow 52x-10=0$
This is not a quadratic equation.
c) Given that, $(k+1)x^2+\frac{3}{2}x=7,wherek=-1$
$\Rightarrow x^2(-1+1)+\frac{3}{2}x=7$
$\Rightarrow 3x-14=0$
Which is also not a quadratic equation.
d) Given that, $x^3-x^2=(x-1{)}^3$
$x^3-x^2=x^3-3x^2\left(1\right)+3x(1{)}^2-(1{)}^3$
$[\therefore (a-b{)}^3=a^3-b^3+3a{b}^2-3a^{2}b]$
$x^3-x^2-x^3+3x^2-3x+1=0$
$2x^2-3x+1=0$
Which represents a quadratic equation because it has the quadratic form.