Question

Consider the quadratic polynomial $16x^2-50x+25$. Which of the following quadratic equations will have roots equal to the zeros of the given quadratic polynomial?

(i) $16x^2-50x=0$

(ii) $4(2x-3{)}^2-(2x-3)-14=0$

(iii) $16x^2-50x+81=25+2x$

Answer 1

$p\left(x\right)=16x^2-50x+25=0$

zeroes of $p\left(x\right)=x=\frac{50\pm \sqrt{2500-1600}}{2\left(16\right)}$

$\Rightarrow x=\frac{5}{2},\frac{5}{8}$

(i)$16x^2-50x=0$

$\Rightarrow x(x-\frac{50}{16})=0$

$\Rightarrow$Not common roots.

Putting (ii)$x=\frac{5}{2}$

$\Rightarrow 4\left(2\right(\frac{5}{2})-3{)}^2-(2\left(\frac{5}{2}\right)-3)-14=16-2-14=0$

$x=\frac{5}{8}$

$\Rightarrow \left(2\right(\frac{5}{8})-3{)}^2-(2\left(\frac{5}{8}\right)-3)-14=16-2-14=0$

$\Rightarrow x=\frac{5}{2},\frac{5}{8}$ are roots of $4(2x-3{)}^2-(2x-3)-14=0$

(iii) $16x^2-50x+81=25+2x$

$16x^2-50x+81-25-2x=0$

$\Rightarrow 16x^2-52x+56=0$

Putting $x=\frac{5}{2}then,16(\frac{5}{2}{)}^2-52(\frac{5}{2})+56\ne 0$

$\Rightarrow$ No common roots.