Solve 4x^2+1=12x

$$4 x ^ { 2 } + 1 = 12 x$$

Solve for x

$$x=\sqrt{2}+\frac{3}{2}\approx 2.914213562$$

$$x=\frac{3}{2}-\sqrt{2}\approx 0.085786438$$

Steps Using the Quadratic Formula

Subtract 12x from both sides.

$$4x^{2}+1-12x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$4x^{2}-12x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -12 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 4}}{2\times 4}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 4}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-12\right)±\sqrt{144-16}}{2\times 4}$$

Add 144 to -16.

$$x=\frac{-\left(-12\right)±\sqrt{128}}{2\times 4}$$

Take the square root of 128.

$$x=\frac{-\left(-12\right)±8\sqrt{2}}{2\times 4}$$

The opposite of -12 is 12.

$$x=\frac{12±8\sqrt{2}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{12±8\sqrt{2}}{8}$$

Now solve the equation x=\frac{12±8\sqrt{2}}{8} when ± is plus. Add 12 to 8\sqrt{2}.

$$x=\frac{8\sqrt{2}+12}{8}$$

Divide 12+8\sqrt{2} by 8.

$$x=\sqrt{2}+\frac{3}{2}$$

Now solve the equation x=\frac{12±8\sqrt{2}}{8} when ± is minus. Subtract 8\sqrt{2} from 12.

$$x=\frac{12-8\sqrt{2}}{8}$$

Divide 12-8\sqrt{2} by 8.

$$x=\frac{3}{2}-\sqrt{2}$$

The equation is now solved.

$$x=\sqrt{2}+\frac{3}{2} x=\frac{3}{2}-\sqrt{2}$$

Steps for Completing the Square

Subtract 12x from both sides.

$$4x^{2}+1-12x=0$$

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

$$4x^{2}-12x=-1$$

Divide both sides by 4.

$$\frac{4x^{2}-12x}{4}=-\frac{1}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{12}{4}\right)x=-\frac{1}{4}$$

Divide -12 by 4.

$$x^{2}-3x=-\frac{1}{4}$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=-\frac{1}{4}+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=\frac{-1+9}{4}$$

Add -\frac{1}{4} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-3x+\frac{9}{4}=2$$

Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{2}\right)^{2}=2$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{2}$$

Simplify.

$$x-\frac{3}{2}=\sqrt{2} x-\frac{3}{2}=-\sqrt{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\sqrt{2}+\frac{3}{2} x=\frac{3}{2}-\sqrt{2}$$