Solve 4x^2-25=0

$$4 x ^ { 2 } - 25 = 0$$

Solve for x

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

Steps Using Difference of Squares

Consider 4x^{2}-25. Rewrite 4x^{2}-25 as \left(2x\right)^{2}-5^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(2x-5\right)\left(2x+5\right)=0$$

To find equation solutions, solve 2x-5=0 and 2x+5=0.

$$x=\frac{5}{2} x=-\frac{5}{2}$$

Steps by Finding Square Root

Add 25 to both sides. Anything plus zero gives itself.

$$4x^{2}=25$$

Divide both sides by 4.

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

Take the square root of both sides of the equation.

$$x=\frac{5}{2} x=-\frac{5}{2}$$

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.

$$4x^{2}-25=0$$

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

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

Square 0.

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

Multiply -4 times 4.

$$x=\frac{0±\sqrt{-16\left(-25\right)}}{2\times 4}$$

Multiply -16 times -25.

$$x=\frac{0±\sqrt{400}}{2\times 4}$$

Take the square root of 400.

$$x=\frac{0±20}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{0±20}{8}$$

Now solve the equation x=\frac{0±20}{8} when ± is plus. Reduce the fraction \frac{20}{8} to lowest terms by extracting and canceling out 4.

$$x=\frac{5}{2}$$

Now solve the equation x=\frac{0±20}{8} when ± is minus. Reduce the fraction \frac{-20}{8} to lowest terms by extracting and canceling out 4.

$$x=-\frac{5}{2}$$

The equation is now solved.

$$x=\frac{5}{2} x=-\frac{5}{2}$$