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How do you solve $2 x \sqrt{5} - \frac{1}{4} > 3 x + \frac{2}{3}$ $2x\sqrt{5}-\frac{1}{4}>3x+\frac{2}{3}$ ?

1 Answer

Jun 20, 2018

$x > \frac{11}{24 \sqrt{5} - 36}$ $x > 11/(24sqrt(5)-36)$

Explanation:

Given
$XXX 2 x \sqrt{5} - \frac{1}{4} > 3 x + \frac{2}{3}$ $color(white)("XXX")2xsqrt(5)-1/4 > 3x+2/3$

Since we can always add or subtract the same amounts from both sides of an inequality
$XXX 2 \sqrt{5} x - 3 x > \frac{2}{3} + \frac{1}{4}$ $color(white)("XXX")2sqrt(5)x-3x > 2/3+1/4$

Simplifying
$XXX (2 \sqrt{5} - 3) x > \frac{8}{12} + \frac{3}{12}$ $color(white)("XXX")(2sqrt(5)-3)x > 8/12+3/12$

and since $(2 \sqrt{5} - 3) > 0$ $(2sqrt(5)-3) > 0$ we can divided both sides by this without changing the direction of the inequality
$XXX x > \frac{\frac{11}{12}}{2 \sqrt{5} - 3}$ $color(white)("XXX")x > (11/12)/(2sqrt(5)-3)$

$XXX x > \frac{11}{24 \sqrt{5} - 36}$ $color(white)("XXX")x > 11/(24sqrt(5)-36)$

If desired, you could use a calculator to approximate the expression on the right side of the inequality above.

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