Question

Find the values of m for which the equation $x^2-15-m(2x-8)=0$ have equal roots.

Answer 1

$x^2-15-m(2x-8)=0$

$x^2-2mx-15+8m=0$

$x^2-2mx-(8m-15)=0$

Here, $a=1,b=-2m,c=8m-15$

The roots of above equation are equal if-

$b^2-4ac=0$

${(-2m)}^2-4(8m-15)\left(1\right)=0$

$4m^2-32m+60=0$

$m^2-8m+15=0$

$m^2-5m-3m+15=0$

$m(m-5)-3(m-5)=0$

$(m-5)(m-3)=0$

$m=5$ or $3$

Hence the value of $m$ for which the given equation have real roots are $3$ and $5$.