Solve 1div2+0.25x^2=1

$$1 \div 2+0.25 { x }^{ 2 } =1$$

Solve for x

$$x=\sqrt{2}\approx 1.414213562$$

$$x=-\sqrt{2}\approx -1.414213562$$

Steps by Finding Square Root

Subtract \frac{1}{2} from both sides.

$$0.25x^{2}=1-\frac{1}{2}$$

Subtract \frac{1}{2} from 1 to get \frac{1}{2}.

$$0.25x^{2}=\frac{1}{2}$$

Divide both sides by 0.25.

$$x^{2}=\frac{\frac{1}{2}}{0.25}$$

Express \frac{\frac{1}{2}}{0.25} as a single fraction.

$$x^{2}=\frac{1}{2\times 0.25}$$

Multiply 2 and 0.25 to get 0.5.

$$x^{2}=\frac{1}{0.5}$$

Expand \frac{1}{0.5} by multiplying both numerator and the denominator by 10.

$$x^{2}=\frac{10}{5}$$

Divide 10 by 5 to get 2.

$$x^{2}=2$$

Take the square root of both sides of the equation.

$$x=\sqrt{2} x=-\sqrt{2}$$

Steps Using the Quadratic Formula

Subtract 1 from both sides.

$$\frac{1}{2}+0.25x^{2}-1=0$$

Subtract 1 from \frac{1}{2} to get -\frac{1}{2}.

$$-\frac{1}{2}+0.25x^{2}=0$$

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.

$$0.25x^{2}-\frac{1}{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 0.25 for a, 0 for b, and -\frac{1}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 0.25\left(-\frac{1}{2}\right)}}{2\times 0.25}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 0.25\left(-\frac{1}{2}\right)}}{2\times 0.25}$$

Multiply -4 times 0.25.

$$x=\frac{0±\sqrt{-\left(-\frac{1}{2}\right)}}{2\times 0.25}$$

Multiply -1 times -\frac{1}{2}.

$$x=\frac{0±\sqrt{\frac{1}{2}}}{2\times 0.25}$$

Take the square root of \frac{1}{2}.

$$x=\frac{0±\frac{\sqrt{2}}{2}}{2\times 0.25}$$

Multiply 2 times 0.25.

$$x=\frac{0±\frac{\sqrt{2}}{2}}{0.5}$$

Now solve the equation x=\frac{0±\frac{\sqrt{2}}{2}}{0.5} when ± is plus.

$$x=\sqrt{2}$$

Now solve the equation x=\frac{0±\frac{\sqrt{2}}{2}}{0.5} when ± is minus.

$$x=-\sqrt{2}$$

The equation is now solved.

$$x=\sqrt{2} x=-\sqrt{2}$$