Question

Find the sum (s) and product (p) of zeros of the polynomial $x^2-3x-5$ by using those value $S$ and $P$ as zeros write another quadratic polynomial ?

Answer 1

Given the polynomial $x^2-3x-5$

For solving we have to make it a zero,

$\therefore x^2-3x-5=0$

Now, by formula,

$x=\frac{-(-3)\pm \sqrt{{(-3)}^2-4\cdot 1\cdot (-5)}}{2\cdot 1}=\frac{3\pm \sqrt{9+20}}{2}=\frac{3\pm \sqrt{29}}{2}\therefore x=\frac{3+\sqrt{29}}{2},\frac{3-\sqrt{29}}{2}$

$\therefore$ Sum of the roots, $\left(S\right)=\frac{3+\sqrt{29}}{2}+\frac{3-\sqrt{29}}{2}=\frac{3+\sqrt{29}+3-\sqrt{29}}{2}=\frac{6}{2}=3$

and product of the roots, $\left(P\right)=\left(\frac{3+\sqrt{29}}{2}\right)\times \left(\frac{3-\sqrt{29}}{2}\right)=\frac{9-29}{4}=\frac{-20}{4}=-5$

Now, using $S$ and $P$ as zeros we have,

$x=3\Rightarrow x-3=0$

and $x=-5\Rightarrow x+5=0$

$\therefore$ the required polynomial is,

$(x-3)(x+5)=x^2+2x-15.$