How do you solve $\sqrt{4 m^{2} - 3 m + 2} - 2 m - 5 = 0$ $sqrt(4m^2-3m+2)-2m-5=0$ ?

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Answer 1 · 2017-02-10T21:29:26

$m = 1$ $m=1$

If we separate out the root and put everything else on the other side of the = we can 'get rid' of the root by squaring both sides.

Write as $\sqrt{4 m^{2} - 3 m + 2} = 2 m + 5$ $sqrt(4m^2-3m+2)=2m+5$

Square both sides

$4 m^{2} - 3 m + 2 = {(2 m + 5)}^{2}$ $4m^2-3m+2" "=" "(2m+5)^2$

$cancel(4m^2)-3m+2" "=" "cancel(4m^2)+20m+25$

$20m+3m=2-25$

$23m=23$

Divide both sides by 23

$m=1$

How do you solve sqrt(4m^2-3m+2)-2m-5=0√4m2−3m+2−2m−5=0sqrt(4m^2-3m+2)-2m-5=0?