Question
Solve $: \left(x^2+5 x+4\right)\left(x^2+5 x+6\right)=120$
✓
Answer
$\left(x^2+5 x+4\right)\left(x^2+5 x+6\right)=120$
Let $x^2+5 x=y$
then $(y + 4)(y + 6) = 120$
$\Rightarrow y^2+6 y+4 y+24-120=0$
$\Rightarrow y^2+10 y-96=0$
$\Rightarrow y^2+16 y-6 y-96=0$
$\Rightarrow y(y+16)-6(y+16)=0$
$\Rightarrow (y+16)(y-16)=0$
then $y = -16$ or $y = 6$
$\Rightarrow x^2+5 x+16=0 \text { or } x^2+5 x-6=0$
$\Rightarrow x=\frac{-5 \pm \sqrt{(5)^2-4(1)(16)}}{2(1)} \text { or } x^2+6 x-x-6=0$
$\Rightarrow x=\frac{-5 \pm \sqrt{-39}}{2} \text { or } x ( x +6)-1( x +6)=0$
$\Rightarrow x=\frac{-5 \pm \sqrt{-39}}{2} \text { or } x ( x +6)-1( x +6)=0$
(reject) or $(x + 6)(x - 1) = 0$
then $x = -6$ and $x = 1$
Let $x^2+5 x=y$
then $(y + 4)(y + 6) = 120$
$\Rightarrow y^2+6 y+4 y+24-120=0$
$\Rightarrow y^2+10 y-96=0$
$\Rightarrow y^2+16 y-6 y-96=0$
$\Rightarrow y(y+16)-6(y+16)=0$
$\Rightarrow (y+16)(y-16)=0$
then $y = -16$ or $y = 6$
$\Rightarrow x^2+5 x+16=0 \text { or } x^2+5 x-6=0$
$\Rightarrow x=\frac{-5 \pm \sqrt{(5)^2-4(1)(16)}}{2(1)} \text { or } x^2+6 x-x-6=0$
$\Rightarrow x=\frac{-5 \pm \sqrt{-39}}{2} \text { or } x ( x +6)-1( x +6)=0$
$\Rightarrow x=\frac{-5 \pm \sqrt{-39}}{2} \text { or } x ( x +6)-1( x +6)=0$
(reject) or $(x + 6)(x - 1) = 0$
then $x = -6$ and $x = 1$
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