Solve 36-4*(a^2+1)5geq0

$$36 - 4 \cdot ( a ^ { 2 } + 1 ) 5 \geq 0$$

Solve for a

$$a\in \begin{bmatrix}-\frac{2\sqrt{5}}{5},\frac{2\sqrt{5}}{5}\end{bmatrix}$$

Solution Steps

Multiply 4 and 5 to get 20.

$$36-20\left(a^{2}+1\right)\geq 0$$

Use the distributive property to multiply -20 by a^{2}+1.

$$36-20a^{2}-20\geq 0$$

Subtract 20 from 36 to get 16.

$$16-20a^{2}\geq 0$$

Multiply the inequality by -1 to make the coefficient of the highest power in 16-20a^{2} positive. Since -1 is negative, the inequality direction is changed.

$$-16+20a^{2}\leq 0$$

Add \frac{4}{5} to both sides.

$$a^{2}\leq \frac{4}{5}$$

Calculate the square root of \frac{4}{5} and get \frac{2\sqrt{5}}{5}. Rewrite \frac{4}{5} as \left(\frac{2\sqrt{5}}{5}\right)^{2}.

$$a^{2}\leq \left(\frac{2\sqrt{5}}{5}\right)^{2}$$

Inequality holds for |a|\leq \frac{2\sqrt{5}}{5}.

$$|a|\leq \frac{2\sqrt{5}}{5}$$

Rewrite |a|\leq \frac{2\sqrt{5}}{5} as a\in \left[-\frac{2\sqrt{5}}{5},\frac{2\sqrt{5}}{5}\right].

$$a\in \begin{bmatrix}-\frac{2\sqrt{5}}{5},\frac{2\sqrt{5}}{5}\end{bmatrix}$$