Question

If the roots of $a^2{x}^2-abx+c=0$ are consecutive integers, then $b^2-a^2=$

• A. $c$
• B. $2c$
• C. $3c$
• D. $4c$

Answer 1

The correct option is D $4c$

Consider the given equation.

$a^2{x}^2-abx+c=0$ $.........\left(1\right)$

Let the roots are $x,x+1$.

Therefore,

$x+x+1=\frac{ab}{a^2}$

$2x+1=\frac{b}{a}$

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

And

$x(x+1)=\frac{c}{a^2}$

$x^2+x=\frac{c}{a^2}$ $...........\left(3\right)$

From equation $\left(2\right)$ and $\left(3\right)$, we get

$(\frac{b-a}{2a}{)}^2+\frac{b-a}{2a}=\frac{c}{a^2}$

$\frac{(b-a{)}^2}{4a^2}+\frac{b-a}{2a}=\frac{c}{a^2}$

$\frac{(b-a{)}^2+2a(b-a)}{4a^2}=\frac{c}{a^2}$

$b^2+a^2-2ab+2ab-2a^2=4c$

$b^2-a^2=4c$

Hence, this is the answer.