Solve :
$(x^{2} - x)^{2} - (x^{2} - x) - 12 = 0$

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Solution

put $x^{2} - x = t$
$t^{2} - t - 12 = 0$
$(t - 4) (t + 3) = 0$
$t = 4, - 3$
$x^{2} - x - 4 = 0$
$x = \frac{1 \pm \sqrt{17}}{2}$
$x^{2} - x + 3 = 0$
$x = \frac{1 \pm \sqrt{10}}{2}$

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Similar questions

x^{2} + x - 12

x^{2} - x - 12 = 0

x^{2} + \frac{x^{2}}{(x + 1)^{2}} = 3

Draw the graph of $y = x^{2} + x - 12$ and hence solve $x^{2} + 2 x + 2 = 0$ .

Solve graphically.

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