Question

If roots of the equation $a{x}^2+bx+c=0are.\frac{\alpha }{\alpha -1}.and.\frac{\alpha +1}{\alpha }$, then prove that $(a+b+c{)}^2=b^2-4ac$.

Answer 1

Given equation : $a{x}^2+bx+c=0⟶\left(1\right)$

roots : $\frac{\alpha }{\alpha -1};\frac{\alpha +1}{\alpha }$

Product of roots $\Rightarrow \frac{\alpha +1}{\alpha -1}=\frac{c}{a}$

$\Rightarrow a\alpha +a=c\alpha -c$

$\Rightarrow \alpha =\frac{-(c+a)}{a-c}$

$\Rightarrow \alpha =\frac{c+a}{c-a}$

substituting value of $\alpha$ in $\alpha /\alpha -1$, we get

$\frac{\alpha }{\alpha -1}=\frac{c+a/c-a}{c+a/c-a-1}$

$=\frac{c+a}{c+a-c+a}=\frac{c+a}{2a}$

substituting value of $\alpha /\alpha -1$ in eq $\left(1\right)$

$\Rightarrow$$a{\left(\frac{c+a}{2a}\right)}^2+b\left(\frac{c+a}{2a}\right)+c=0$

$\Rightarrow$$a(c^2+a^2=2ac)+2abc+2a^{2}b+4a^{2}c=0$

$\Rightarrow$$c^2+a^2+2ac+2bc+2ab+4ac=0$

Adding $b^2$ on both sides,

$\Rightarrow$$a^2+b^2+c^2+2ab+2bc+2ac=b^2-4ac$

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

Hence proved.