Question

One root of the quadratic equation $a{x}^2+bx+c=0$, is cube of the other. Then prove that $ca(c+a{)}^2=(b^2-2ac{)}^2$.

Answer 1

Let the roots of the equation $a{x}^2+bx+c=0$ be $\alpha ,{\alpha }^3$.

Then from the relation between the roots and coefficient we get,

$\alpha +{\alpha }^3=-\frac{b}{a}$......(1) and ${\alpha }^4=\frac{c}{a}$......(2).

From (1) we get,

$\alpha (1+{\alpha }^2)=-\frac{b}{a}$

Now squaring both sides we get,

or, ${\alpha }^2(1+{\alpha }^2{)}^2=\frac{b^2}{a^2}$

or, ${\alpha }^2(1+2{\alpha }^2+{\alpha }^4)=\frac{b^2}{a^2}$

or, ${\alpha }^2(1+{\alpha }^4)=\frac{b^2}{a^2}-2{\alpha }^4$

or, ${\alpha }^2(1+\frac{c}{a})=\frac{b^2}{a^2}-2\frac{c}{a}$ [ Using (2)]

Again squaring both sides we get,

or, ${\alpha }^4{(1+\frac{c}{a})}^2={(\frac{b^2}{a^2}-2\frac{c}{a})}^2$

or, $\frac{c}{a}{(1+\frac{c}{a})}^2={(\frac{b^2}{a^2}-2\frac{c}{a})}^2$

or, $ac(a+c{)}^2=(b^2-2ac{)}^2$.