Question
Solve :
$(x^2 - 3x)^2 - 16(x^2 - 3x) - 36 =0$
$(x^2 - 3x)^2 - 16(x^2 - 3x) - 36 =0$
✓
Answer
$(x^2 - 3x)^2 - 16(x^2 - 3x) - 36 =0$ Let $x^2 - 3x = y$
then $y^2 - 16y - 36 = 0$
$\Rightarrow y^2 - 18y + 2y - 36 = 0$
$\Rightarrow y(y - 18) + 2(y - 18) = 0$
$\Rightarrow ( y - 18) (y + 2) = 0$
If $y -18=0$ or $y +2=0$
$\Rightarrow x^2 - 3x - 18 = 0$ or $x^2 - 3x + 2 = 0$
$\Rightarrow x^2 - 6x + 3x - 18 = 0 or x^2 -2x x + 2 = 0$
$\Rightarrow x( x -6) + 3(x - 6) = 0 or x( x -2) -1 (x -2) = 0$
$\Rightarrow (x -6) ( x + 3) = 0 or (x -2)(x -1) = 0$
If $x-6=0$ or $x+3=0$ or $x-2=0$ or $x-1=0$
then $x=6$ or $x=-3$ or $x=2$ or $x=1$
then $y^2 - 16y - 36 = 0$
$\Rightarrow y^2 - 18y + 2y - 36 = 0$
$\Rightarrow y(y - 18) + 2(y - 18) = 0$
$\Rightarrow ( y - 18) (y + 2) = 0$
If $y -18=0$ or $y +2=0$
$\Rightarrow x^2 - 3x - 18 = 0$ or $x^2 - 3x + 2 = 0$
$\Rightarrow x^2 - 6x + 3x - 18 = 0 or x^2 -2x x + 2 = 0$
$\Rightarrow x( x -6) + 3(x - 6) = 0 or x( x -2) -1 (x -2) = 0$
$\Rightarrow (x -6) ( x + 3) = 0 or (x -2)(x -1) = 0$
If $x-6=0$ or $x+3=0$ or $x-2=0$ or $x-1=0$
then $x=6$ or $x=-3$ or $x=2$ or $x=1$
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