How do you evaluate square root of 4/3 minus the square root of 3/4?

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How do you evaluate square root of 4/3 minus the square root of 3/4?

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Answer 1 · 2018-04-09T05:11:24

Please look below.

Explanation:

Let $a = \sqrt{\frac{4}{3}} - \sqrt{\frac{3}{4}}$ $a = sqrt(4/3) - sqrt(3/4)$
Let $b = \sqrt{\frac{4}{3}} + \sqrt{\frac{3}{4}}$ $b = sqrt(4/3) + sqrt(3/4)$

$a b = \frac{4}{3} - \frac{3}{4}$ $ab = 4/3 - 3/4$
$a b = \frac{7}{12}$ $ab = 7/12$
$a + b = 2 \sqrt{\frac{4}{3}}$ $a+b = 2sqrt(4/3)$
$a + b = \frac{4}{\sqrt{3}}$ $a+b = 4/sqrt3$

Now as simultaneous equations:
$a (\frac{4}{\sqrt{3}} - a) = \frac{7}{12}$ $a(4/sqrt3 -a) = 7/12$
$a^{2} - \frac{4 a}{\sqrt{3}} + \frac{7}{12} = 0$ $a^2 - (4a)/sqrt3 + 7/12 = 0$

Quadratic equation:
$a = \frac{\frac{4}{\sqrt{3}} \pm \sqrt{{(\frac{4}{\sqrt{3}})}^{2} - 4 \times \frac{7}{12}}}{2}$ $a = (4/sqrt3 +- sqrt((4/sqrt3)^2 - 4 xx 7/12))/(2)$

$a = \frac{\frac{4}{\sqrt{3}} \pm \sqrt{\frac{16}{3} - \frac{7}{3}}}{2}$ $a = (4/sqrt3 +-sqrt(16/3 - 7/3))/(2)$

$a = \frac{\frac{4}{\sqrt{3}} \pm \frac{3}{\sqrt{3}}}{2}$ $a = (4/sqrt(3) +- 3/sqrt3)/2$

$a = \frac{\frac{4}{\sqrt{3}} - \frac{3}{\sqrt{3}}}{2}$ $a = (4/sqrt(3) - 3/sqrt3)/2$

$a = \frac{1}{2 \sqrt{3}}$ $a = 1/(2sqrt3)$

Therefore $\sqrt{\frac{4}{3}} - \sqrt{\frac{3}{4}} = \frac{1}{2 \sqrt{3}}$ $sqrt(4/3) - sqrt(3/4) = 1/(2sqrt3)$
and we also get $\sqrt{\frac{4}{3}} + \sqrt{\frac{3}{4}} = \frac{7}{2 \sqrt{3}}$ $sqrt(4/3) + sqrt(3/4) = 7/(2sqrt3)$

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How do you evaluate square root of 4/3 minus the square root of 3/4?

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