Question
Solve : $(x^2 – x)^2 + 5(x^2 – x)+ 4=0$
✓
Answer
$\left(x^2-x\right)^2+5\left(x^2-x\right)+4=0$
Let $x^2-x=y$
Then $y^2+5 y+4=0$
$\Rightarrow y^2+4 y+y+4=0 $
$ \Rightarrow y(y+4)+1(y+4)=0$
$\Rightarrow(y+4)(y+1)=0$
if $y+4=0$ or $y+1=0$
$\Rightarrow x^2-x+4=0$ or $x^2-x+1=0$
$\Rightarrow x=\frac{-(-1) \pm \sqrt{(-1)^2-4(1)(4)}}{2(1)}$ or $\frac{-(-1) \pm \sqrt{(-1)^2-4(1)(1)}}{2(1)}$
$\Rightarrow x=\frac{1 \pm \sqrt{-15}}{2}$ (reject) or $x=\frac{1 \pm \sqrt{-3}}{2}$ (reject)
$\therefore$ Given equation has no real solution
Let $x^2-x=y$
Then $y^2+5 y+4=0$
$\Rightarrow y^2+4 y+y+4=0 $
$ \Rightarrow y(y+4)+1(y+4)=0$
$\Rightarrow(y+4)(y+1)=0$
if $y+4=0$ or $y+1=0$
$\Rightarrow x^2-x+4=0$ or $x^2-x+1=0$
$\Rightarrow x=\frac{-(-1) \pm \sqrt{(-1)^2-4(1)(4)}}{2(1)}$ or $\frac{-(-1) \pm \sqrt{(-1)^2-4(1)(1)}}{2(1)}$
$\Rightarrow x=\frac{1 \pm \sqrt{-15}}{2}$ (reject) or $x=\frac{1 \pm \sqrt{-3}}{2}$ (reject)
$\therefore$ Given equation has no real solution
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