Solve 25x^2-81=0

$$25 x ^ { 2 } - 81 = 0$$

Solve for x

$$x = \frac{9}{5} = 1\frac{4}{5} = 1.8$$

$$x = -\frac{9}{5} = -1\frac{4}{5} = -1.8$$

Steps Using Difference of Squares

Consider 25x^{2}-81. Rewrite 25x^{2}-81 as \left(5x\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).

$$\left(5x-9\right)\left(5x+9\right)=0$$

To find equation solutions, solve 5x-9=0 and 5x+9=0.

$$x=\frac{9}{5} x=-\frac{9}{5}$$

Steps by Finding Square Root

Add 81 to both sides. Anything plus zero gives itself.

$$25x^{2}=81$$

Divide both sides by 25.

$$x^{2}=\frac{81}{25}$$

Take the square root of both sides of the equation.

$$x=\frac{9}{5} x=-\frac{9}{5}$$

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.

$$25x^{2}-81=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, 0 for b, and -81 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 25\left(-81\right)}}{2\times 25}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 25\left(-81\right)}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{0±\sqrt{-100\left(-81\right)}}{2\times 25}$$

Multiply -100 times -81.

$$x=\frac{0±\sqrt{8100}}{2\times 25}$$

Take the square root of 8100.

$$x=\frac{0±90}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{0±90}{50}$$

Now solve the equation x=\frac{0±90}{50} when ± is plus. Reduce the fraction \frac{90}{50} to lowest terms by extracting and canceling out 10.

$$x=\frac{9}{5}$$

Now solve the equation x=\frac{0±90}{50} when ± is minus. Reduce the fraction \frac{-90}{50} to lowest terms by extracting and canceling out 10.

$$x=-\frac{9}{5}$$

The equation is now solved.

$$x=\frac{9}{5} x=-\frac{9}{5}$$