How do you solve $sqrt(m+2)<sqrt(3m+4)$ ?

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Answer 1 · 2017-08-14T01:47:37

$m > -1$ .

If we square both sides, we can get:

$(sqrt(m + 2))^2 < (sqrt(3m + 4))^2$

$m + 2 < 3m + 4$

$-2m < 2$

We now divide both sides by $-2$ , not forgetting to switch the direction of the inequality symbol.

$m > -1$

We now test our answer to see if it is true. Let $m = 1$ .

$sqrt(3) < sqrt(7) color(green)(√)$

This is obviously true.

Hopefully this helps!

How do you solve sqrt(m+2)<sqrt(3m+4)sqrt(m+2)<sqrt(3m+4)?