Solve (23-x)^2+x^2=17^2

$$( 23 - x ) ^ { 2 } + x ^ { 2 } = 17 ^ { 2 }$$

Solve for x

$$x=8$$

$$x=15$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(23-x\right)^{2}.

$$529-46x+x^{2}+x^{2}=17^{2}$$

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

$$529-46x+2x^{2}=17^{2}$$

Calculate 17 to the power of 2 and get 289.

$$529-46x+2x^{2}=289$$

Subtract 289 from both sides.

$$529-46x+2x^{2}-289=0$$

Subtract 289 from 529 to get 240.

$$240-46x+2x^{2}=0$$

Divide both sides by 2.

$$120-23x+x^{2}=0$$

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

$$x^{2}-23x+120=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+120. To find a and b, set up a system to be solved.

$$a+b=-23 ab=1\times 120=120$$

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 120.

$$-1,-120 -2,-60 -3,-40 -4,-30 -5,-24 -6,-20 -8,-15 -10,-12$$

Calculate the sum for each pair.

$$-1-120=-121 -2-60=-62 -3-40=-43 -4-30=-34 -5-24=-29 -6-20=-26 -8-15=-23 -10-12=-22$$

The solution is the pair that gives sum -23.

$$a=-15 b=-8$$

Rewrite x^{2}-23x+120 as \left(x^{2}-15x\right)+\left(-8x+120\right).

$$\left(x^{2}-15x\right)+\left(-8x+120\right)$$

Factor out x in the first and -8 in the second group.

$$x\left(x-15\right)-8\left(x-15\right)$$

Factor out common term x-15 by using distributive property.

$$\left(x-15\right)\left(x-8\right)$$

To find equation solutions, solve x-15=0 and x-8=0.

$$x=15 x=8$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(23-x\right)^{2}.

$$529-46x+x^{2}+x^{2}=17^{2}$$

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

$$529-46x+2x^{2}=17^{2}$$

Calculate 17 to the power of 2 and get 289.

$$529-46x+2x^{2}=289$$

Subtract 289 from both sides.

$$529-46x+2x^{2}-289=0$$

Subtract 289 from 529 to get 240.

$$240-46x+2x^{2}=0$$

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.

$$2x^{2}-46x+240=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -46 for b, and 240 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-46\right)±\sqrt{\left(-46\right)^{2}-4\times 2\times 240}}{2\times 2}$$

Square -46.

$$x=\frac{-\left(-46\right)±\sqrt{2116-4\times 2\times 240}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-46\right)±\sqrt{2116-8\times 240}}{2\times 2}$$

Multiply -8 times 240.

$$x=\frac{-\left(-46\right)±\sqrt{2116-1920}}{2\times 2}$$

Add 2116 to -1920.

$$x=\frac{-\left(-46\right)±\sqrt{196}}{2\times 2}$$

Take the square root of 196.

$$x=\frac{-\left(-46\right)±14}{2\times 2}$$

The opposite of -46 is 46.

$$x=\frac{46±14}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{46±14}{4}$$

Now solve the equation x=\frac{46±14}{4} when ± is plus. Add 46 to 14.

$$x=\frac{60}{4}$$

Divide 60 by 4.

$$x=15$$

Now solve the equation x=\frac{46±14}{4} when ± is minus. Subtract 14 from 46.

$$x=\frac{32}{4}$$

Divide 32 by 4.

$$x=8$$

The equation is now solved.

$$x=15 x=8$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(23-x\right)^{2}.

$$529-46x+x^{2}+x^{2}=17^{2}$$

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

$$529-46x+2x^{2}=17^{2}$$

Calculate 17 to the power of 2 and get 289.

$$529-46x+2x^{2}=289$$

Subtract 529 from both sides.

$$-46x+2x^{2}=289-529$$

Subtract 529 from 289 to get -240.

$$-46x+2x^{2}=-240$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$2x^{2}-46x=-240$$

Divide both sides by 2.

$$\frac{2x^{2}-46x}{2}=-\frac{240}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{46}{2}\right)x=-\frac{240}{2}$$

Divide -46 by 2.

$$x^{2}-23x=-\frac{240}{2}$$

Divide -240 by 2.

$$x^{2}-23x=-120$$

Divide -23, the coefficient of the x term, by 2 to get -\frac{23}{2}. Then add the square of -\frac{23}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-23x+\left(-\frac{23}{2}\right)^{2}=-120+\left(-\frac{23}{2}\right)^{2}$$

Square -\frac{23}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-23x+\frac{529}{4}=-120+\frac{529}{4}$$

Add -120 to \frac{529}{4}.

$$x^{2}-23x+\frac{529}{4}=\frac{49}{4}$$

Factor x^{2}-23x+\frac{529}{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}.

$$\left(x-\frac{23}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{23}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x-\frac{23}{2}=\frac{7}{2} x-\frac{23}{2}=-\frac{7}{2}$$

Add \frac{23}{2} to both sides of the equation.

$$x=15 x=8$$