Question

Solve the following quadratic equation.

${\displaystyle \frac{1}{4-\text{p}}-\frac{1}{2+\text{p}}=\frac{1}{4}}$

Answer 1

${\displaystyle \frac{1}{4-\text{p}}-\frac{1}{2+\text{p}}=\frac{1}{4}}$

∴ ${\displaystyle \frac{2+\text{p}-(4-\text{p})}{(4-\text{p})(2+\text{p})}=\frac{1}{4}}$

∴ ${\displaystyle \frac{2+\text{p}-4+\text{p}}{8+4\text{p}-2\text{p}-{\text{p}}^2}=\frac{1}{4}}$

∴ ${\displaystyle \frac{2\text{p}-2}{8+2\text{p}-{\text{p}}^2}=\frac{1}{4}}$

∴ 4(2p – 2) = 8 + 2p – p²

∴ 8p – 8 = 8 + 2p – p²

∴ p² – 2p + 8p – 8 – 8 = 0

∴ p² + 6p – 16 = 0

-16

8  -2

8 × (-2) = -16

8 - 2 = 6

∴ p² + 8p – 2p – 16 = 0

∴ p(p + 8) – 2(p + 8) = 0

∴ (p + 8)(p – 2) = 0

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get

p + 8 = 0 or p – 2 = 0

∴ p = – 8 or p = 2

∴ The roots of the given equation are – 8 and 2.