Question

(x² + x)(x² + x – 7) + 10 = 0

Answer 1

Given: (x² + x)(x² + x – 7) + 10 = 0
Let x² + x = m. Then,
m(m – 7) + 10 = 0
$\Rightarrow$ m² – 7m + 10 = 0
On splitting the middle term –7m as –5m – 2m, we get:
m² – 5m – 2m + 10 = 0
$\Rightarrow$ m(m – 5) – 2(m – 5) = 0
$\Rightarrow$ (m – 5)(m – 2) = 0
$\Rightarrow$ m – 2 = 0 or m – 5 = 0
$\Rightarrow$ m = 2 or m = 5
On substituting m = x² + x, we get
$\Rightarrow$ x² + x = 2 or x² + x = 5
$\Rightarrow$ x² + x – 2 = 0 …(1) or x² + x – 5 = 0 …(2)
Now, from equation (1), we get:
x² + x – 2 = 0
$\Rightarrow$ x² + 2x – x – 2 = 0
$\Rightarrow$ x(x + 2) – 1(x + 2) = 0
$\Rightarrow$ (x + 2)(x – 1) = 0
$\Rightarrow$ x + 2 = 0 or x – 1 = 0
$\Rightarrow$ x = –2 or x = 1

Now, from equation (2), we get:
x² + x – 5 = 0
On using the quadratic formula, we get:
$$\begin{gathered} x=\frac{-1\pm \sqrt{{\left(1\right)}^2-4\times 1\times \left(-5\right)}}{2\times 1} \\ =\frac{-1\pm \sqrt{1+20}}{2} \\ =\frac{-1\pm \sqrt{21}}{2} \end{gathered}$$
Thus, the solutions of the given equation are x = $\frac{-1\pm \sqrt{21}}{2}$ , –2, 1.