Solve 25left(1-x^2right)=16

$$25 \left( 1- { x }^{ 2 } \right) = 16$$

Solve for x

$$x=\frac{3}{5}=0.6$$

$$x=-\frac{3}{5}=-0.6$$

Steps by Finding Square Root

Divide both sides by 25.

$$1-x^{2}=\frac{16}{25}$$

Subtract 1 from both sides.

$$-x^{2}=\frac{16}{25}-1$$

Subtract 1 from \frac{16}{25} to get -\frac{9}{25}.

$$-x^{2}=-\frac{9}{25}$$

Divide both sides by -1.

$$x^{2}=\frac{-\frac{9}{25}}{-1}$$

Express \frac{-\frac{9}{25}}{-1} as a single fraction.

$$x^{2}=\frac{-9}{25\left(-1\right)}$$

Multiply 25 and -1 to get -25.

$$x^{2}=\frac{-9}{-25}$$

Fraction \frac{-9}{-25} can be simplified to \frac{9}{25} by removing the negative sign from both the numerator and the denominator.

$$x^{2}=\frac{9}{25}$$

Take the square root of both sides of the equation.

$$x=\frac{3}{5} x=-\frac{3}{5}$$

Steps Using the Quadratic Formula

Divide both sides by 25.

$$1-x^{2}=\frac{16}{25}$$

Subtract \frac{16}{25} from both sides.

$$1-x^{2}-\frac{16}{25}=0$$

Subtract \frac{16}{25} from 1 to get \frac{9}{25}.

$$\frac{9}{25}-x^{2}=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$-x^{2}+\frac{9}{25}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 0 for b, and \frac{9}{25} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-1\right)\times \frac{9}{25}}}{2\left(-1\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-1\right)\times \frac{9}{25}}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{0±\sqrt{4\times \frac{9}{25}}}{2\left(-1\right)}$$

Multiply 4 times \frac{9}{25}.

$$x=\frac{0±\sqrt{\frac{36}{25}}}{2\left(-1\right)}$$

Take the square root of \frac{36}{25}.

$$x=\frac{0±\frac{6}{5}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{0±\frac{6}{5}}{-2}$$

Now solve the equation x=\frac{0±\frac{6}{5}}{-2} when ± is plus.

$$x=-\frac{3}{5}$$

Now solve the equation x=\frac{0±\frac{6}{5}}{-2} when ± is minus.

$$x=\frac{3}{5}$$

The equation is now solved.

$$x=-\frac{3}{5} x=\frac{3}{5}$$