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

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Answer 1 · 2017-08-14T01:47:37

$m > - 1$ $m > -1$ .

If we square both sides, we can get:

${(\sqrt{m + 2})}^{2} < {(\sqrt{3 m + 4})}^{2}$ $(sqrt(m + 2))^2 < (sqrt(3m + 4))^2$

$m + 2 < 3 m + 4$ $m + 2 < 3m + 4$

$- 2 m < 2$ $-2m < 2$

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

$m > - 1$ $m > -1$

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

$\sqrt{3} < \sqrt{7} \sqrt$ $sqrt(3) < sqrt(7) color(green)(√)$

This is obviously true.

Hopefully this helps!

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