Question

p² – 12p + 32 = 0

Answer 1

Given: p² – 12p + 32 = 0

$\Rightarrow$ p² – 12p = –32 … (1)

We make L.H.S. a perfect square by using

Third term = ${\left(\frac{1}{2}\times \mathrm{coefficient}\mathrm{of}p\right)}^2={\left(\frac{1}{2}\times \left(-12\right)\right)}^2={\left(-6\right)}^2=36$

Adding 36 to both sides of equation (1), we get:

p²– 12p + 36 = –32 + 36

$\Rightarrow$p² – 2 × p × 6 + (6)2 = 4

$\Rightarrow$(p – 6)² = (2)²

On taking square root of both sides, we get:

p – 6 = $\pm 2$

$\Rightarrow$p = 6 ± 2
= 6 + 2, 6 – 2
= 8, 4