Solve x^2-81/256=0

$$x ^ { 2 } - \frac { 81 } { 256 } = 0$$

Solve for x

$$x=\frac{9}{16}=0.5625$$

$$x=-\frac{9}{16}=-0.5625$$

Steps Using Difference of Squares

Multiply both sides by 256.

$$256x^{2}-81=0$$

Consider 256x^{2}-81. Rewrite 256x^{2}-81 as \left(16x\right)^{2}-9^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(16x-9\right)\left(16x+9\right)=0$$

To find equation solutions, solve 16x-9=0 and 16x+9=0.

$$x=\frac{9}{16} x=-\frac{9}{16}$$

Steps by Finding Square Root

Add \frac{81}{256} to both sides. Anything plus zero gives itself.

$$x^{2}=\frac{81}{256}$$

Take the square root of both sides of the equation.

$$x=\frac{9}{16} x=-\frac{9}{16}$$

Steps Using the Quadratic Formula

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.

$$x^{2}-\frac{81}{256}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -\frac{81}{256} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-\frac{81}{256}\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-\frac{81}{256}\right)}}{2}$$

Multiply -4 times -\frac{81}{256}.

$$x=\frac{0±\sqrt{\frac{81}{64}}}{2}$$

Take the square root of \frac{81}{64}.

$$x=\frac{0±\frac{9}{8}}{2}$$

Now solve the equation x=\frac{0±\frac{9}{8}}{2} when ± is plus.

$$x=\frac{9}{16}$$

Now solve the equation x=\frac{0±\frac{9}{8}}{2} when ± is minus.

$$x=-\frac{9}{16}$$

The equation is now solved.

$$x=\frac{9}{16} x=-\frac{9}{16}$$