Solve 4x^2={left(6-1/2xright)}^2+3/4x^2

$$4 { x }^{ 2 } = { \left(6- \frac{ 1 }{ 2 } x \right) }^{ 2 } + \frac{ 3 }{ 4 } { x }^{ 2 }$$

Solve for x

$$x=\sqrt{13}-1\approx 2.605551275$$

$$x=-\left(\sqrt{13}+1\right)\approx -4.605551275$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6-\frac{1}{2}x\right)^{2}.

$$4x^{2}=36-6x+\frac{1}{4}x^{2}+\frac{3}{4}x^{2}$$

Combine \frac{1}{4}x^{2} and \frac{3}{4}x^{2} to get x^{2}.

$$4x^{2}=36-6x+x^{2}$$

Subtract 36 from both sides.

$$4x^{2}-36=-6x+x^{2}$$

Add 6x to both sides.

$$4x^{2}-36+6x=x^{2}$$

Subtract x^{2} from both sides.

$$4x^{2}-36+6x-x^{2}=0$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-36+6x=0$$

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.

$$3x^{2}+6x-36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 6 for b, and -36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 3\left(-36\right)}}{2\times 3}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 3\left(-36\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-6±\sqrt{36-12\left(-36\right)}}{2\times 3}$$

Multiply -12 times -36.

$$x=\frac{-6±\sqrt{36+432}}{2\times 3}$$

Add 36 to 432.

$$x=\frac{-6±\sqrt{468}}{2\times 3}$$

Take the square root of 468.

$$x=\frac{-6±6\sqrt{13}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-6±6\sqrt{13}}{6}$$

Now solve the equation x=\frac{-6±6\sqrt{13}}{6} when ± is plus. Add -6 to 6\sqrt{13}.

$$x=\frac{6\sqrt{13}-6}{6}$$

Divide -6+6\sqrt{13} by 6.

$$x=\sqrt{13}-1$$

Now solve the equation x=\frac{-6±6\sqrt{13}}{6} when ± is minus. Subtract 6\sqrt{13} from -6.

$$x=\frac{-6\sqrt{13}-6}{6}$$

Divide -6-6\sqrt{13} by 6.

$$x=-\sqrt{13}-1$$

The equation is now solved.

$$x=\sqrt{13}-1 x=-\sqrt{13}-1$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6-\frac{1}{2}x\right)^{2}.

$$4x^{2}=36-6x+\frac{1}{4}x^{2}+\frac{3}{4}x^{2}$$

Combine \frac{1}{4}x^{2} and \frac{3}{4}x^{2} to get x^{2}.

$$4x^{2}=36-6x+x^{2}$$

Add 6x to both sides.

$$4x^{2}+6x=36+x^{2}$$

Subtract x^{2} from both sides.

$$4x^{2}+6x-x^{2}=36$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}+6x=36$$

Divide both sides by 3.

$$\frac{3x^{2}+6x}{3}=\frac{36}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{6}{3}x=\frac{36}{3}$$

Divide 6 by 3.

$$x^{2}+2x=\frac{36}{3}$$

Divide 36 by 3.

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

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square 1.

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

Add 12 to 1.

$$x^{2}+2x+1=13$$

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+1\right)^{2}=13$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{13}$$

Simplify.

$$x+1=\sqrt{13} x+1=-\sqrt{13}$$

Subtract 1 from both sides of the equation.

$$x=\sqrt{13}-1 x=-\sqrt{13}-1$$