Solve 2x^2+32=16x

$$2 x ^ { 2 } + 32 = 16 x$$

Solve for x

$$x=4$$

Steps Using Factoring By Grouping

Subtract 16x from both sides.

$$2x^{2}+32-16x=0$$

Divide both sides by 2.

$$x^{2}+16-8x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}-8x+16=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+16. To find a and b, set up a system to be solved.

$$a+b=-8 ab=1\times 16=16$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 16.

$$-1,-16 -2,-8 -4,-4$$

Calculate the sum for each pair.

$$-1-16=-17 -2-8=-10 -4-4=-8$$

The solution is the pair that gives sum -8.

$$a=-4 b=-4$$

Rewrite x^{2}-8x+16 as \left(x^{2}-4x\right)+\left(-4x+16\right).

$$\left(x^{2}-4x\right)+\left(-4x+16\right)$$

Factor out x in the first and -4 in the second group.

$$x\left(x-4\right)-4\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(x-4\right)$$

Rewrite as a binomial square.

$$\left(x-4\right)^{2}$$

To find equation solution, solve x-4=0.

$$x=4$$

Steps Using the Quadratic Formula

Subtract 16x from both sides.

$$2x^{2}+32-16x=0$$

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.

$$2x^{2}-16x+32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -16 for b, and 32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 2\times 32}}{2\times 2}$$

Square -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\times 2\times 32}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-16\right)±\sqrt{256-8\times 32}}{2\times 2}$$

Multiply -8 times 32.

$$x=\frac{-\left(-16\right)±\sqrt{256-256}}{2\times 2}$$

Add 256 to -256.

$$x=\frac{-\left(-16\right)±\sqrt{0}}{2\times 2}$$

Take the square root of 0.

$$x=-\frac{-16}{2\times 2}$$

The opposite of -16 is 16.

$$x=\frac{16}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{16}{4}$$

Divide 16 by 4.

$$x=4$$

Steps for Completing the Square

Subtract 16x from both sides.

$$2x^{2}+32-16x=0$$

Subtract 32 from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}-16x=-32$$

Divide both sides by 2.

$$\frac{2x^{2}-16x}{2}=-\frac{32}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{16}{2}\right)x=-\frac{32}{2}$$

Divide -16 by 2.

$$x^{2}-8x=-\frac{32}{2}$$

Divide -32 by 2.

$$x^{2}-8x=-16$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=-16+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=-16+16$$

Add -16 to 16.

$$x^{2}-8x+16=0$$

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-4\right)^{2}=0$$

Take the square root of both sides of the equation.

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

Simplify.

$$x-4=0 x-4=0$$

Add 4 to both sides of the equation.

$$x=4 x=4$$

The equation is now solved. Solutions are the same.

$$x=4$$