Solve 25/4x^2-9/4=0

$$\frac { 25 } { 4 } x ^ { 2 } - \frac { 9 } { 4 } = 0$$

Solve for x

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

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

Steps Using Difference of Squares

Multiply both sides by 4.

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

Consider 25x^{2}-9. Rewrite 25x^{2}-9 as \left(5x\right)^{2}-3^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(5x-3\right)\left(5x+3\right)=0$$

To find equation solutions, solve 5x-3=0 and 5x+3=0.

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

Steps by Finding Square Root

Add \frac{9}{4} to both sides. Anything plus zero gives itself.

$$\frac{25}{4}x^{2}=\frac{9}{4}$$

Multiply both sides by \frac{4}{25}, the reciprocal of \frac{25}{4}.

$$x^{2}=\frac{9}{4}\times \frac{4}{25}$$

Multiply \frac{9}{4} and \frac{4}{25} to get \frac{9}{25}.

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

Take the square root of both sides of the equation.

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

Steps Using the Quadratic Formula

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.

$$\frac{25}{4}x^{2}-\frac{9}{4}=0$$

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

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

Square 0.

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

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

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

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

$$x=\frac{0±\sqrt{\frac{225}{4}}}{2\times \frac{25}{4}}$$

Take the square root of \frac{225}{4}.

$$x=\frac{0±\frac{15}{2}}{2\times \frac{25}{4}}$$

Multiply 2 times \frac{25}{4}.

$$x=\frac{0±\frac{15}{2}}{\frac{25}{2}}$$

Now solve the equation x=\frac{0±\frac{15}{2}}{\frac{25}{2}} when ± is plus.

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

Now solve the equation x=\frac{0±\frac{15}{2}}{\frac{25}{2}} when ± is minus.

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

The equation is now solved.

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