How do you solve $x^{2} - 4 x \geq 5$ $x^2-4x>=5$ ?

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Answer 1 · 2018-02-01T19:15:39

$x^{2} - 4 x \geq 5$ $x^2−4x≥5$

$x^{2} - 4 x - 5 \geq 0$ $x^2−4x-5≥0$

$x^{2} + x - 5 x - 5 \geq 0$ $x^2+x-5x-5≥0$

$x (x + 1) - 5 (x + 1) \geq 0$ $x(x+1)-5(x+1)≥0$

$(x - 5) (x + 1) \geq 0$ $(x-5)(x+1)≥0$

$:.(x-5)≥0$ and $(x+1)≥0$

$x>=5$ and $x>=-1$

$x<=5$ as it will also be greater than -1 then.

-Sahar.

How do you solve x^2-4x>=5x2−4x≥5x^2-4x>=5?