How do you solve $x^{2} - 4 x \geq 0$ $x^2-4x>=0$ using a sign chart?

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Answer 1 · 2017-10-15T09:45:10

We first consider where $x^{2} - 4 x = 0$ $x^2 - 4x = 0$

i.e. when $x (x - 4) = 0$ $x(x - 4) = 0$

Hence, $x = 0 or x = 4$ $x = 0 or x = 4$

We then need to consider three seperate regions.

i.e. $x < 0$ $x < 0$ , $0 < x < 4$ $0 < x < 4$ and $x > 4$ $x > 4$

For $x < 0$ $x < 0$ , we have that $x^{2} - 4 x > 0$ $x^2 - 4x > 0$

For $0 < x < 4$ $0 < x < 4$ , we have that $x^{2} - 4 x < 0$ $x^2 - 4x < 0$

and for $x > 4$ $x > 4$ we have that $x^{2} - 4 x > 0$ $x^2 - 4x > 0$

Hence, for $x^{2} - 4 x \geq 0$ $x^2 - 4x >=0$ we have two regions.

i.e. $x \leq 0$ $x <=0$ or $x \geq 4$ $x >=4$

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How do you solve x2−4x≥0x^2-4x>=0 using a sign chart?