Question
Solve for $x: (x^2 - 5x)^2 - 7(x^2 - 5x) + 6 = 0; x \in R.$
✓
Answer
Given equation
$\left(x^2-5 x\right)^2-7\left(x^2-5 x\right)+6=0$
Put $x^2-5 x=y$
$\therefore$ The given equation becomes
$y^2-7 y+6=0$
$\Rightarrow y^2-6 y-y+6=0$
$\Rightarrow y(y-6)-1(y-6)=0$
$\Rightarrow y=1,6$
$\text { But } x^2-5 x=y$
$\therefore x^2-5 x=1$
$x^2-5 x-1=0$
$\text { Here } a=1, b=-5, c=-1$
$\therefore x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} $
$ x=\frac{-(-5) \pm \sqrt{25+4}}{2} $
$x=\frac{5 \pm \sqrt{29}}{2}$
$x^2-5 x=6 \\ \Rightarrow x^2-5 x-6=0 $
$ \Rightarrow x^2-6 x+x-6=0$
$ \Rightarrow x(x-6)+1(x-6)=0$
$ \Rightarrow(x-6)(x+1)=0 $
$\Rightarrow x=6 \text { or } x=-1$
Hence, the roots are $-1, 6,$ $\frac{5 \pm \sqrt{29}}{2}$.
$\left(x^2-5 x\right)^2-7\left(x^2-5 x\right)+6=0$
Put $x^2-5 x=y$
$\therefore$ The given equation becomes
$y^2-7 y+6=0$
$\Rightarrow y^2-6 y-y+6=0$
$\Rightarrow y(y-6)-1(y-6)=0$
$\Rightarrow y=1,6$
$\text { But } x^2-5 x=y$
$\therefore x^2-5 x=1$
$x^2-5 x-1=0$
$\text { Here } a=1, b=-5, c=-1$
$\therefore x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} $
$ x=\frac{-(-5) \pm \sqrt{25+4}}{2} $
$x=\frac{5 \pm \sqrt{29}}{2}$
$x^2-5 x=6 \\ \Rightarrow x^2-5 x-6=0 $
$ \Rightarrow x^2-6 x+x-6=0$
$ \Rightarrow x(x-6)+1(x-6)=0$
$ \Rightarrow(x-6)(x+1)=0 $
$\Rightarrow x=6 \text { or } x=-1$
Hence, the roots are $-1, 6,$ $\frac{5 \pm \sqrt{29}}{2}$.
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