Question

Which of the following values of $p$ satisfy the equation?

$(p^2+p)(p^2+p-3)=28$

• A. $p=\frac{1\pm \sqrt{29}}{2}$
• B. $p=\frac{-1\pm \sqrt{15}}{2}$
• C. $p=\frac{-1\pm \sqrt{29}}{2}$
• D. $p=\frac{1\pm \sqrt{15}}{2}$

Answer 1

Answer: (C)

$p=\frac{-1\pm \sqrt{29}}{2}$

$(p^2+p)(p^2+p-3)=28$
Let $p^2+p=x$
Hence,
$x(x-3)=28$
$x^2-3x-28=0$
$x^2-7x+4x-28=0$
$(x-7)(x+4)=0$
$x=7,-4$
Now, when $x=7$, $p^2+p=7$
$p^2+p-7=0$
$p=\frac{-1\pm \sqrt{29}}{2}$
when $x=-4$, $p^2+p=-4$
$p^2+p+4=0$

$p=\frac{-1\pm \sqrt{1-16}}{2}$
so not possible to find real roots when p=$-4$.
Thus, $p=\frac{-1\pm \sqrt{29}}{2}$