# How do you solve 25-x^2>=0?

Apr 10, 2018

Solution: $- 5 \le x \le 5 \mathmr{and} \left[- 5 , 5\right]$

#### Explanation:

$25 - {x}^{2} \ge 0 \mathmr{and} \left(5 + x\right) \left(5 - x\right) \ge 0$

Critical points are

$5 + x = 0 \mathmr{and} x = - 5 \mathmr{and} 5 - x = 0 \mathmr{and} x = 5$

$f \left(x\right) = 0$ when $x = - 5 \mathmr{and} x = 5$

Sign chart:

When $x < - 5$ sign of $\left(5 + x\right) \left(5 - x\right)$ is  (-) * (+) = (-) ; < 0

When $- 5 < x < 5$ sign of $\left(5 + x\right) \left(5 - x\right)$ is  (+) * (+) = (+) ; > 0

When $x > 5$ sign of $\left(5 + x\right) \left(5 - x\right)$ is  (+) * (-) = (-) ; < 0

Solution: $- 5 \le x \le 5 \mathmr{and} \left[- 5 , 5\right]$ [Ans]

Apr 10, 2018

color(blue)([ -5 ,5]

#### Explanation:

$25 - {x}^{2} \ge 0$

Factor $L H S$:

$- \left(5 + x\right) \left(5 - x\right) \ge 0$

$\left(5 + x\right) \left(5 - x\right) \le 0$

For:

$5 + x \le 0$

$x \le - 5$

For:

$5 - x \le 0$

$x \ge 5$

Solutions:

color(blue)([ -5 ,5]