Question

p² – 10p + 5 = 0

Answer 1

Given: p² – 10p + 5 = 0
=> p² ¬ 10p = –5 … (1)
We make L.H.S. a perfect square by using
Third term =$={\left(\frac{1}{2}\times \mathrm{coefficient}\mathrm{of}\mathrm{p}\right)}^2={\left(\frac{1}{2}\times (-10)\right)}^2=(-5{)}^2=25$
On adding 25 to both sides of equation (1), we get:
$$\begin{gathered} p^2-10p+25=-5+25=20 \\ =>p^2-2\times p\times 5+(5{)}^2={\left(2\sqrt{5}\right)}^2 \\ =>(p-5{)}^2={\left(2\sqrt{5}\right)}^2 \\ \mathrm{On}\mathrm{taking}\mathrm{square}\mathrm{root}\mathrm{of}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}: \\ p-5=\pm 2\sqrt{5} \\ =>p=5\pm 2\sqrt{5} \\ \mathrm{Thus},p=5\pm 2\sqrt{5} \end{gathered}$$