Solve quad(2)4x^2+1=4x*x=

$$\quad ( 2 ) 4 x ^ { 2 } + 1 = 4 x \cdot x =$$

Solve for x (complex solution)

$$x=-\frac{1}{2}i=-0.5i$$

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

Steps by Finding Square Root

Multiply x and x to get x^{2}.

$$2\times 4x^{2}+1=4x^{2}$$

Multiply 2 and 4 to get 8.

$$8x^{2}+1=4x^{2}$$

Subtract 4x^{2} from both sides.

$$8x^{2}+1-4x^{2}=0$$

Combine 8x^{2} and -4x^{2} to get 4x^{2}.

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

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 4.

$$x^{2}=-\frac{1}{4}$$

The equation is now solved.

$$x=\frac{1}{2}i x=-\frac{1}{2}i$$

Steps Using the Quadratic Formula

Multiply x and x to get x^{2}.

$$2\times 4x^{2}+1=4x^{2}$$

Multiply 2 and 4 to get 8.

$$8x^{2}+1=4x^{2}$$

Subtract 4x^{2} from both sides.

$$8x^{2}+1-4x^{2}=0$$

Combine 8x^{2} and -4x^{2} to get 4x^{2}.

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

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

$$x=\frac{0±\sqrt{0^{2}-4\times 4}}{2\times 4}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 4}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{0±\sqrt{-16}}{2\times 4}$$

Take the square root of -16.

$$x=\frac{0±4i}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{0±4i}{8}$$

Now solve the equation x=\frac{0±4i}{8} when ± is plus.

$$x=\frac{1}{2}i$$

Now solve the equation x=\frac{0±4i}{8} when ± is minus.

$$x=-\frac{1}{2}i$$

The equation is now solved.

$$x=\frac{1}{2}i x=-\frac{1}{2}i$$