question_answer

If the quadratic equation \[p{{x}^{2}}-2\sqrt{5}\,px+15=0,\] has two equal roots then find the value of p.

Answer:

The given quadratic equation is,

\[p{{x}^{2}}-2\sqrt{5}\,px+15=0,\]

This is of the form

\[a{{x}^{2}}+bx+c=0\]

Where, \[a=p,b=-2\sqrt{5}p,c=15\]

We have,                       \[D={{b}^{2}}-4ac\]

\[={{(-2\sqrt{5}p)}^{2}}-4\times p\times 15\]

\[=20{{p}^{2}}-60p\]

\[=20p(p-3)\]

For real and equal roots/ we must have:

\[D=0,\,\,\,\Rightarrow 20p(p-3)=0\]

\[\Rightarrow p=0,p=3\]

\[p=0\], is not possible as whole equation will be zero. Hence, 3 is the required value of p.