Solve (3-x)^2+x^2=5

$$( 3 - x ) ^ { 2 } + x ^ { 2 } = 5$$

Solve for x

$$x=1$$

$$x=2$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-x\right)^{2}.

$$9-6x+x^{2}+x^{2}=5$$

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

$$9-6x+2x^{2}=5$$

Subtract 5 from both sides.

$$9-6x+2x^{2}-5=0$$

Subtract 5 from 9 to get 4.

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

Divide both sides by 2.

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

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

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

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

$$a+b=-3 ab=1\times 2=2$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-2 b=-1$$

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

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

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

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

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

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

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

$$x=2 x=1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-x\right)^{2}.

$$9-6x+x^{2}+x^{2}=5$$

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

$$9-6x+2x^{2}=5$$

Subtract 5 from both sides.

$$9-6x+2x^{2}-5=0$$

Subtract 5 from 9 to get 4.

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

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.

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

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

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

Square -6.

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

Multiply -4 times 2.

$$x=\frac{-\left(-6\right)±\sqrt{36-8\times 4}}{2\times 2}$$

Multiply -8 times 4.

$$x=\frac{-\left(-6\right)±\sqrt{36-32}}{2\times 2}$$

Add 36 to -32.

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

Take the square root of 4.

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

The opposite of -6 is 6.

$$x=\frac{6±2}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{6±2}{4}$$

Now solve the equation x=\frac{6±2}{4} when ± is plus. Add 6 to 2.

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

Divide 8 by 4.

$$x=2$$

Now solve the equation x=\frac{6±2}{4} when ± is minus. Subtract 2 from 6.

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

Divide 4 by 4.

$$x=1$$

The equation is now solved.

$$x=2 x=1$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-x\right)^{2}.

$$9-6x+x^{2}+x^{2}=5$$

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

$$9-6x+2x^{2}=5$$

Subtract 9 from both sides.

$$-6x+2x^{2}=5-9$$

Subtract 9 from 5 to get -4.

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

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.

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

Divide both sides by 2.

$$\frac{2x^{2}-6x}{2}=-\frac{4}{2}$$

Dividing by 2 undoes the multiplication by 2.

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

Divide -6 by 2.

$$x^{2}-3x=-\frac{4}{2}$$

Divide -4 by 2.

$$x^{2}-3x=-2$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=-2+\frac{9}{4}$$

Add -2 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{1}{4}$$

Factor x^{2}-3x+\frac{9}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-\frac{3}{2}=\frac{1}{2} x-\frac{3}{2}=-\frac{1}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=2 x=1$$