Solve x^2+(7-x)^2=25

$$x ^ { 2 } + ( 7 - x ) ^ { 2 } = 25$$

Solve for x

$$x=3$$

$$x=4$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(7-x\right)^{2}.

$$x^{2}+49-14x+x^{2}=25$$

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

$$2x^{2}+49-14x=25$$

Subtract 25 from both sides.

$$2x^{2}+49-14x-25=0$$

Subtract 25 from 49 to get 24.

$$2x^{2}+24-14x=0$$

Divide both sides by 2.

$$x^{2}+12-7x=0$$

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

$$x^{2}-7x+12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=-7 ab=1\times 12=12$$

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 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -7.

$$a=-4 b=-3$$

Rewrite x^{2}-7x+12 as \left(x^{2}-4x\right)+\left(-3x+12\right).

$$\left(x^{2}-4x\right)+\left(-3x+12\right)$$

Factor out x in the first and -3 in the second group.

$$x\left(x-4\right)-3\left(x-4\right)$$

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

$$\left(x-4\right)\left(x-3\right)$$

To find equation solutions, solve x-4=0 and x-3=0.

$$x=4 x=3$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(7-x\right)^{2}.

$$x^{2}+49-14x+x^{2}=25$$

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

$$2x^{2}+49-14x=25$$

Subtract 25 from both sides.

$$2x^{2}+49-14x-25=0$$

Subtract 25 from 49 to get 24.

$$2x^{2}+24-14x=0$$

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.

$$2x^{2}-14x+24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -14 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 2\times 24}}{2\times 2}$$

Square -14.

$$x=\frac{-\left(-14\right)±\sqrt{196-4\times 2\times 24}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-14\right)±\sqrt{196-8\times 24}}{2\times 2}$$

Multiply -8 times 24.

$$x=\frac{-\left(-14\right)±\sqrt{196-192}}{2\times 2}$$

Add 196 to -192.

$$x=\frac{-\left(-14\right)±\sqrt{4}}{2\times 2}$$

Take the square root of 4.

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

The opposite of -14 is 14.

$$x=\frac{14±2}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{14±2}{4}$$

Now solve the equation x=\frac{14±2}{4} when ± is plus. Add 14 to 2.

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

Divide 16 by 4.

$$x=4$$

Now solve the equation x=\frac{14±2}{4} when ± is minus. Subtract 2 from 14.

$$x=\frac{12}{4}$$

Divide 12 by 4.

$$x=3$$

The equation is now solved.

$$x=4 x=3$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(7-x\right)^{2}.

$$x^{2}+49-14x+x^{2}=25$$

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

$$2x^{2}+49-14x=25$$

Subtract 49 from both sides.

$$2x^{2}-14x=25-49$$

Subtract 49 from 25 to get -24.

$$2x^{2}-14x=-24$$

Divide both sides by 2.

$$\frac{2x^{2}-14x}{2}=-\frac{24}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{14}{2}\right)x=-\frac{24}{2}$$

Divide -14 by 2.

$$x^{2}-7x=-\frac{24}{2}$$

Divide -24 by 2.

$$x^{2}-7x=-12$$

Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=-12+\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=-12+\frac{49}{4}$$

Add -12 to \frac{49}{4}.

$$x^{2}-7x+\frac{49}{4}=\frac{1}{4}$$

Factor x^{2}-7x+\frac{49}{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}.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{1}{2} x-\frac{7}{2}=-\frac{1}{2}$$

Add \frac{7}{2} to both sides of the equation.

$$x=4 x=3$$