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.
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