How do you solve $4 \sqrt{3 m^{2} - 15} = 4$ $4sqrt(3m^2-15)=4$ ?

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Answer 1 · 2017-02-20T21:15:59

$m = \sqrt{\frac{16}{3}}$ $m=sqrt(16/3)$ and $- \sqrt{\frac{16}{3}}$ $-sqrt(16/3)$

Divide 4 on both sides:

$(\frac{4}{4}) \sqrt{3 m^{2} - 15} = (\frac{4}{4})$ $(4/4)sqrt(3m^2-15)=(4/4)$

The fours cancel leaving:

$\sqrt{3 m^{2} - 15} = 1$ $sqrt(3m^2-15)=1$

Square both sides to get rid of the square root. Note that when you do this, it will become positive or negative 1

$3 m^{2} - 15 = 1$ $3m^2-15=1$

Add 15 to both sides.

$3 m^{2} = 16$ $3m^2=16$

Divide 3 on both sides.

$m^{2} = \frac{16}{3}$ $m^2=16/3$

Finally, square root both sides to get rid of the square. Note that this will give you the positive and the negative of the answer

$m = \sqrt{\frac{16}{3}}$ $m=sqrt(16/3)$ and $- \sqrt{\frac{16}{3}}$ $-sqrt(16/3)$

How do you solve 4√3m2−15=44sqrt(3m^2-15)=4?