Solve 4x^2+1=5

$$4 { x }^{ 2 } +1 = 5$$

Solve for x

$$x=1$$

$$x=-1$$

Steps Using Difference of Squares

Subtract 5 from both sides.

$$4x^{2}+1-5=0$$

Subtract 5 from 1 to get -4.

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

Divide both sides by 4.

$$x^{2}-1=0$$

Consider x^{2}-1. Rewrite x^{2}-1 as x^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-1\right)\left(x+1\right)=0$$

To find equation solutions, solve x-1=0 and x+1=0.

$$x=1 x=-1$$

Steps by Finding Square Root

Subtract 1 from both sides.

$$4x^{2}=5-1$$

Subtract 1 from 5 to get 4.

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

Divide both sides by 4.

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

Divide 4 by 4 to get 1.

$$x^{2}=1$$

Take the square root of both sides of the equation.

$$x=1 x=-1$$

Steps Using the Quadratic Formula

Subtract 5 from both sides.

$$4x^{2}+1-5=0$$

Subtract 5 from 1 to get -4.

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

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

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

Square 0.

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

Multiply -4 times 4.

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

Multiply -16 times -4.

$$x=\frac{0±\sqrt{64}}{2\times 4}$$

Take the square root of 64.

$$x=\frac{0±8}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{0±8}{8}$$

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

$$x=1$$

Now solve the equation x=\frac{0±8}{8} when ± is minus. Divide -8 by 8.

$$x=-1$$

The equation is now solved.

$$x=1 x=-1$$