Question

Find the quadratic polynomial, sum of whose zeros is 8 and their product is 12. Hence, find the zeros of the polynomial.

Answer 1

$\begin{array}{l}\text{Let}\alpha \text{and}\beta \text{be the zeroes of the required polynomial}f\left(x\right).\\ \text{Then (}\alpha \text{+}\beta \text{)=8 and}\alpha \beta \text{=12}\\ \therefore f\left(x\right)=x^2-\text{(}\alpha \text{+}\beta \text{)}x+\alpha \beta \\ =>f\left(x\right)=x^2-8x+12\\ \text{Hence, required polynomial}f\left(x\right)=x^2-8x+12\\ \therefore f\left(x\right)=0=>x^2-8x+12=0\\ \text{=>}{x}^2-(6x+2x)+12=0\\ \text{=>}{x}^2-6x-2x+12=0\\ \text{=>}x(x-6)-2(x-6)=0\\ \text{=> (}x-\text{2)}(x-6)=0\\ \text{=> (}x-\text{2)}=0\text{or}(x-6)=0\\ \text{=>}x=2\text{or}x=6\\ \text{So, the zeroes of}f\left(x\right)\text{are}2\text{and}6.\end{array}$