• # question_answer If $5$ is a root of the quadratic equation $2{{x}^{2}}+px-15=0$ and the quadratic equation $p({{x}^{2}}+x)k=0$ has equal roots, find the value of k.

 Given, $5$ is a root of $2{{x}^{2}}+px-15=0$ then,     $f(-5)=2{{(-5)}^{2}}+p(-5)-15=0$ $50-5p-15=0$ $35-5p=0$ $5p=35$ $p=7$ Now, putting the value of p, in, $p({{x}^{2}}+x)+k=0$ we get               $7{{x}^{2}}+7x+k=0$ Now,                 $D={{b}^{2}}-4ac=0$ ($\because$ has the equal roots) then,                 $49-28k=0$ $28k=49$ $k=\frac{49}{28}=\frac{7}{4}$