Solve 25(x^2+1)=26

$$25( { x }^{ 2 } +1)=26$$

Solve for x

$$x=\frac{1}{5}=0.2$$

$$x=-\frac{1}{5}=-0.2$$

Steps Using Difference of Squares

Divide both sides by 25.

$$x^{2}+1=\frac{26}{25}$$

Subtract \frac{26}{25} from both sides.

$$x^{2}+1-\frac{26}{25}=0$$

Subtract \frac{26}{25} from 1 to get -\frac{1}{25}.

$$x^{2}-\frac{1}{25}=0$$

Multiply both sides by 25.

$$25x^{2}-1=0$$

Consider 25x^{2}-1. Rewrite 25x^{2}-1 as \left(5x\right)^{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).

$$\left(5x-1\right)\left(5x+1\right)=0$$

To find equation solutions, solve 5x-1=0 and 5x+1=0.

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

Steps by Finding Square Root

Divide both sides by 25.

$$x^{2}+1=\frac{26}{25}$$

Subtract 1 from both sides.

$$x^{2}=\frac{26}{25}-1$$

Subtract 1 from \frac{26}{25} to get \frac{1}{25}.

$$x^{2}=\frac{1}{25}$$

Take the square root of both sides of the equation.

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

Steps Using the Quadratic Formula

Divide both sides by 25.

$$x^{2}+1=\frac{26}{25}$$

Subtract \frac{26}{25} from both sides.

$$x^{2}+1-\frac{26}{25}=0$$

Subtract \frac{26}{25} from 1 to get -\frac{1}{25}.

$$x^{2}-\frac{1}{25}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -\frac{1}{25} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-\frac{1}{25}\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-\frac{1}{25}\right)}}{2}$$

Multiply -4 times -\frac{1}{25}.

$$x=\frac{0±\sqrt{\frac{4}{25}}}{2}$$

Take the square root of \frac{4}{25}.

$$x=\frac{0±\frac{2}{5}}{2}$$

Now solve the equation x=\frac{0±\frac{2}{5}}{2} when ± is plus.

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

Now solve the equation x=\frac{0±\frac{2}{5}}{2} when ± is minus.

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

The equation is now solved.

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