Question

Solve: $x(x+5)(x+7)(x+12)+150=0$

• A. $\{6\pm \sqrt{31},6\pm \sqrt{6}\}$
• B. $\{-6\pm \sqrt{31},-6\pm \sqrt{6}\}$
• C. $\{-6\pm \sqrt{6},-6\pm \sqrt{31}\}$
• D. None of these

Answer 1

Answer: (B)

$\{-6\pm \sqrt{31},-6\pm \sqrt{6}\}$

Given $x(x+5)(x+7)(x+12)+150=0$

$\Rightarrow x(x+12)(x+5)(x+7)+150=0$

$\Rightarrow (x^2+12x)(x^2+12x+35)+150=0$

Let $x^2+12x=a$

$a(a+35)+150=0$

$\Rightarrow a^2+35a+150=0$

$\Rightarrow a^2+30a+5a+150=0$

$\Rightarrow (a+30)(a+5)=0$

$\Rightarrow a=-30,-5$

If $x^2+12x=-30$

$\Rightarrow x^2+12x+30=0$

Thus, $x=\frac{-12\pm \sqrt{144-4\times 30}}{2}$ $[\text{Using}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}]$

$=\frac{-12\pm \sqrt{24}}{2}$

$=-6\pm \sqrt{6}$

If $x^2+12=5$

$\Rightarrow x^2+12x+5=0$

$x=\frac{-12\pm \sqrt{144-20}}{2}$

$=-6\pm \sqrt{31}$

Hence, $x=-6\pm \sqrt{31},-6\pm \sqrt{6}$