Question

If one root of the equation $a{x}^2+bx+c=0$ is a square of the other other, then $a(b-c{)}^3=cX$, where $X$ is

• A. $a^3-b^3$
• B. $a^3+b^3$
• C. $(a-b{)}^3$
• D. $(b-a{)}^3$

Answer 1

The correct option is D $(b-a{)}^3$

Let $p$ and $p^2$ be the roots of the equation $a{x}^2+bx+c=0$

$p+p^2=-b/a$

$p(1+p)=-b/a$ ......... (i)

$p\left(p^2\right)=c/a$ ...... (ii)

dividing eq. (i) by (ii)

$(1+p)/p^2=-b/c$

$c(1+p)=-b{p}^2$

$b{p}^2+cp+c=0$ .... (iii)

Also, $a{p}^2+bp+c=0$ ....... (iv)

Subtracting eq. (iv) from (iii), we get

$p^2(b-a)+p(c-b)=0$

$p(b-a)+(c-b)=0$

$p=(b-c)/(b-a)$ .....(v)

Now, from (v) and (ii)

$(b-c{)}^3/(b-a{)}^3=c/a$

$a(b-c{)}^3=c(b-a{)}^3$

now comparing with, $a(b-c{)}^3=cX$, we get

$X=(b-a{)}^3$