Question

If the roots of $a{x}^2+bx+c=0$ are $2$, $\frac{3}{2}$ then ${\left(a+b+c\right)}^2=$

Answer 1

$a{x}^2+bx+c=0$

roots are $(2,\frac{3}{2})$

$\Rightarrow$ $(x-2)(x-\frac{3}{2})=a{x}^2+bx+c$

$\Rightarrow$ $x^2-(2+\frac{3}{2})x+2\times \frac{3}{2}=a{x}^2+bx+c$

$\Rightarrow$ $a=1;b=-(2+\frac{3}{2});c=2$

$a=1;b=-\left(\frac{7}{2}\right);c=2$

${(a+b+c)}^2={(1-\frac{7}{2}+2)}^2$

$={\left(\frac{6-7}{2}\right)}^2=\frac{1}{4}$

$\therefore$ ${(a+b+c)}^2=\frac{1}{4}$