Mathematics - Quadratic Equation Question with Solution | TestHub

Since $P\left(1\right)=1+b+c$ and $P\left(2\right)=4+2b+c$ are both roots of P , we have, from Viet's that $P(1)+P(2)=-b\Rightarrow 4b+2c=-5$ and $(1+b+c)\cdot (4+2b+c)=c$

substituting $b=-\frac{2c+5}{4}$,

we get $\left(-\frac{1}{4}+\frac{c}{2}\right)\cdot \left(\frac{3}{2}\right)=c\Rightarrow \frac{4c-2}{8}\cdot \frac{3}{2}=c\Rightarrow c=-\frac{3}{2}$