How do you solve the inequality $9 \geq 4 - 2 \sqrt{5 x + 7}$ $9>= 4 - 2sqrt( 5x+7)$ ?

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Answer 1 · 2016-02-03T05:01:04

Treat the inequality sign like an equal sign.

9 $\geq 4 - 2 \sqrt{5 x + 7}$ $>= 4 - 2sqrt(5x + 7)$

5 $\geq - 2 \sqrt{5 x + 7}$ $>=- 2sqrt(5x + 7)$

You will have to switch the in equation sign around since you're dividing by a negative number.

$- \frac{5}{2} \leq \sqrt{5 x + 7}$ $-5/2 <= sqrt(5x + 7)$

${(- \frac{5}{2})}^{2} \leq {(\sqrt{5 x + 7})}^{2}$ $(-5/2)^2 <= (sqrt(5x + 7))^2$

$\frac{25}{4} \leq 5 x + 7$ $25/4 <= 5x + 7$

$\frac{25}{4} - \frac{28}{4} \leq 5 x$ $25/4 - 28/4 <= 5x$

$- \frac{3}{4} \times \frac{1}{5} \leq x$ $-3/4 xx 1/5 <= x$

$- \frac{3}{20} \leq x$ $-3/20 <= x$

So, your solution is $- \frac{3}{20} \leq x$ $-3/20 <= x$

Practice Exercises:

a) $\sqrt{3 x + 6} > 2$ $sqrt(3x + 6) > 2$

b) $\sqrt{3 x + 4} + \sqrt{8 x + 4} \leq 10$ $sqrt(3x + 4) + sqrt(8x + 4) <= 10$

Good luck!

How do you solve the inequality 9≥4−2√5x+79>= 4 - 2sqrt( 5x+7) ?