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{\frac{\left(x+2\right)\left(x+2\right)}{6}}\geq {\frac{x}{3}+\frac{1}{2}}

Solve the inequality ((x+2)(x+2))/6%x/3+1/2

Answer

$x\geq \sqrt{6\left(\frac{x}{3}+\frac{1}{2}\right)-\left(4+4x\right)}$

Step-by-step explanation

Problem to solve:

${\frac{\left(x+2\right)\left(x+2\right)}{6}}\geq {\frac{x}{3}+\frac{1}{2}}$
1

When multiplying exponents with same base you can add the exponents

$\frac{\left(x+2\right)^2}{6}\geq \frac{x}{3}+\frac{1}{2}$
2

Moving the $6$ multiplying to the other side of the inequation

$\left(x+2\right)^2\geq 6\left(\frac{x}{3}+\frac{1}{2}\right)$

Unlock this step-by-step solution!

Answer

$x\geq \sqrt{6\left(\frac{x}{3}+\frac{1}{2}\right)-\left(4+4x\right)}$
${\frac{\left(x+2\right)\left(x+2\right)}{6}}\geq {\frac{x}{3}+\frac{1}{2}}$

Main topic:

Inequalities

Time to solve it:

~ 0.96 seconds