Question

Find the sets of values of x for which x(x+2) < x.​

Answer 1

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      Hi my lil bunny!

❧⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯☙

Let's solve your inequality step-by-step.

$x(x+2)<x$

$x^2 + 2x < x$

Let's find the critical points of the inequality.

$x^2 + 2x = x$

$x^2 + 2x - x - x$  (Subtract x from both sides)

$x^2 + x = 0$

$x ( x + 1 ) = 0$ (Factor left side of equation)

$x = 0 |OR| x + 1 = 0$ (Set factors equal to 0)

$x = 0 |OR| x = -1$

Check intervals in between critical points. (Test values in the intervals to see if they work.)

$x < - 1$ (Doesn't work in original inequality)

$-1 < x < 0$ (Works in original inequality)

$x > 0$  (Doesn't work in original inequality)

So the answer is : $-1 < x < 0$

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If this helped you, could you maybe give brainliest..?

❀*May*❀