Solve x^2+left(5-xright)^2-16=2x(5-x)

$${ x }^{ 2 } + { \left(5-x \right) }^{ 2 } -16=2x(5-x)$$

Solve for x

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

$$x = \frac{9}{2} = 4\frac{1}{2} = 4.5$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-x\right)^{2}.

$$x^{2}+25-10x+x^{2}-16=2x\left(5-x\right)$$

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

$$2x^{2}+25-10x-16=2x\left(5-x\right)$$

Subtract 16 from 25 to get 9.

$$2x^{2}+9-10x=2x\left(5-x\right)$$

Use the distributive property to multiply 2x by 5-x.

$$2x^{2}+9-10x=10x-2x^{2}$$

Subtract 10x from both sides.

$$2x^{2}+9-10x-10x=-2x^{2}$$

Combine -10x and -10x to get -20x.

$$2x^{2}+9-20x=-2x^{2}$$

Add 2x^{2} to both sides.

$$2x^{2}+9-20x+2x^{2}=0$$

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

$$4x^{2}+9-20x=0$$

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

$$4x^{2}-20x+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx+9. To find a and b, set up a system to be solved.

$$a+b=-20 ab=4\times 9=36$$

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 36.

$$-1,-36 -2,-18 -3,-12 -4,-9 -6,-6$$

Calculate the sum for each pair.

$$-1-36=-37 -2-18=-20 -3-12=-15 -4-9=-13 -6-6=-12$$

The solution is the pair that gives sum -20.

$$a=-18 b=-2$$

Rewrite 4x^{2}-20x+9 as \left(4x^{2}-18x\right)+\left(-2x+9\right).

$$\left(4x^{2}-18x\right)+\left(-2x+9\right)$$

Factor out 2x in the first and -1 in the second group.

$$2x\left(2x-9\right)-\left(2x-9\right)$$

Factor out common term 2x-9 by using distributive property.

$$\left(2x-9\right)\left(2x-1\right)$$

To find equation solutions, solve 2x-9=0 and 2x-1=0.

$$x=\frac{9}{2} x=\frac{1}{2}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-x\right)^{2}.

$$x^{2}+25-10x+x^{2}-16=2x\left(5-x\right)$$

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

$$2x^{2}+25-10x-16=2x\left(5-x\right)$$

Subtract 16 from 25 to get 9.

$$2x^{2}+9-10x=2x\left(5-x\right)$$

Use the distributive property to multiply 2x by 5-x.

$$2x^{2}+9-10x=10x-2x^{2}$$

Subtract 10x from both sides.

$$2x^{2}+9-10x-10x=-2x^{2}$$

Combine -10x and -10x to get -20x.

$$2x^{2}+9-20x=-2x^{2}$$

Add 2x^{2} to both sides.

$$2x^{2}+9-20x+2x^{2}=0$$

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

$$4x^{2}+9-20x=0$$

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.

$$4x^{2}-20x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -20 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 4\times 9}}{2\times 4}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 4\times 9}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-20\right)±\sqrt{400-16\times 9}}{2\times 4}$$

Multiply -16 times 9.

$$x=\frac{-\left(-20\right)±\sqrt{400-144}}{2\times 4}$$

Add 400 to -144.

$$x=\frac{-\left(-20\right)±\sqrt{256}}{2\times 4}$$

Take the square root of 256.

$$x=\frac{-\left(-20\right)±16}{2\times 4}$$

The opposite of -20 is 20.

$$x=\frac{20±16}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{20±16}{8}$$

Now solve the equation x=\frac{20±16}{8} when ± is plus. Add 20 to 16.

$$x=\frac{36}{8}$$

Reduce the fraction \frac{36}{8} to lowest terms by extracting and canceling out 4.

$$x=\frac{9}{2}$$

Now solve the equation x=\frac{20±16}{8} when ± is minus. Subtract 16 from 20.

$$x=\frac{4}{8}$$

Reduce the fraction \frac{4}{8} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

$$x=\frac{9}{2} x=\frac{1}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-x\right)^{2}.

$$x^{2}+25-10x+x^{2}-16=2x\left(5-x\right)$$

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

$$2x^{2}+25-10x-16=2x\left(5-x\right)$$

Subtract 16 from 25 to get 9.

$$2x^{2}+9-10x=2x\left(5-x\right)$$

Use the distributive property to multiply 2x by 5-x.

$$2x^{2}+9-10x=10x-2x^{2}$$

Subtract 10x from both sides.

$$2x^{2}+9-10x-10x=-2x^{2}$$

Combine -10x and -10x to get -20x.

$$2x^{2}+9-20x=-2x^{2}$$

Add 2x^{2} to both sides.

$$2x^{2}+9-20x+2x^{2}=0$$

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

$$4x^{2}+9-20x=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$4x^{2}-20x=-9$$

Divide both sides by 4.

$$\frac{4x^{2}-20x}{4}=-\frac{9}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{20}{4}\right)x=-\frac{9}{4}$$

Divide -20 by 4.

$$x^{2}-5x=-\frac{9}{4}$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-\frac{9}{4}+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=\frac{-9+25}{4}$$

Add -\frac{9}{4} to \frac{25}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-5x+\frac{25}{4}=4$$

Factor x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{4}$$

Simplify.

$$x-\frac{5}{2}=2 x-\frac{5}{2}=-2$$

Add \frac{5}{2} to both sides of the equation.

$$x=\frac{9}{2} x=\frac{1}{2}$$