Question

(p² + p)(p² + p – 3) = 28

Answer 1

Given: (p² + p)(p² + p – 3) = 28
Let (p² + p) = m
Then, the equation can be written as:
m(m – 3) = 28
=> m² – 3m – 28 = 0

On splitting the middle term –3m as 4m – 7m, we get:

m² + 4m – 7m – 28 = 0
=> m(m + 4) – 7(m + 4) = 0
=> (m + 4)(m – 7) = 0
=> m + 4 = 0 or m – 7 = 0
=> m = – 4 or m = 7

On substituting m = p² + p, we get:

p² + p = –4 or p² + p = 7
=> p² + p + 4 = 0…(1) or p² + p – 7 = 0…(2)
Now, from equation (1), we get:
p² + p+ 4 = 0
On using the quadratic formula, we get:

$$\begin{gathered} p=\frac{-1\pm \sqrt{(1{)}^2-4(1\left)\right(4)}}{2\left(1\right)} \\ =\frac{-1\pm \sqrt{1-16}}{2}=\frac{-1\pm \sqrt{-15}}{2} \end{gathered}$$

^{$$\begin{gathered} Now,\mathrm{from}\mathrm{equation}\left(2\right),\mathrm{we}\mathrm{get}: \\ {p}^2+p-7=0 \\ \text{On using the quadratic formula,we get:} \\ p=\frac{-1\pm \sqrt{{\left(1\right)}^2-4\times 1\times (-7)}}{2\times 1} \\ =\frac{-1\pm \sqrt{1+28}}{2} \\ =\frac{-1\pm \sqrt{29}}{2} \end{gathered}$$}