Solve (x-2)^2+(x+1)^2=x+7

$$( x - 2 ) ^ { 2 } + ( x + 1 ) ^ { 2 } = x + 7$$

Solve for x

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

$$x=2$$

Steps Using Factoring By Grouping

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

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

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

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

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

Combine -4x and 2x to get -2x.

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

Add 4 and 1 to get 5.

$$2x^{2}-2x+5=x+7$$

Subtract x from both sides.

$$2x^{2}-2x+5-x=7$$

Combine -2x and -x to get -3x.

$$2x^{2}-3x+5=7$$

Subtract 7 from both sides.

$$2x^{2}-3x+5-7=0$$

Subtract 7 from 5 to get -2.

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

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

$$a+b=-3 ab=2\left(-2\right)=-4$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -4.

$$1,-4 2,-2$$

Calculate the sum for each pair.

$$1-4=-3 2-2=0$$

The solution is the pair that gives sum -3.

$$a=-4 b=1$$

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

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

Factor out 2x in 2x^{2}-4x.

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

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

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

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

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

Steps Using the Quadratic Formula

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

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

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

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

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

Combine -4x and 2x to get -2x.

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

Add 4 and 1 to get 5.

$$2x^{2}-2x+5=x+7$$

Subtract x from both sides.

$$2x^{2}-2x+5-x=7$$

Combine -2x and -x to get -3x.

$$2x^{2}-3x+5=7$$

Subtract 7 from both sides.

$$2x^{2}-3x+5-7=0$$

Subtract 7 from 5 to get -2.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -3 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 2\left(-2\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-3\right)±\sqrt{9-8\left(-2\right)}}{2\times 2}$$

Multiply -8 times -2.

$$x=\frac{-\left(-3\right)±\sqrt{9+16}}{2\times 2}$$

Add 9 to 16.

$$x=\frac{-\left(-3\right)±\sqrt{25}}{2\times 2}$$

Take the square root of 25.

$$x=\frac{-\left(-3\right)±5}{2\times 2}$$

The opposite of -3 is 3.

$$x=\frac{3±5}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{3±5}{4}$$

Now solve the equation x=\frac{3±5}{4} when ± is plus. Add 3 to 5.

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

Divide 8 by 4.

$$x=2$$

Now solve the equation x=\frac{3±5}{4} when ± is minus. Subtract 5 from 3.

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

Reduce the fraction \frac{-2}{4} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Steps for Completing the Square

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

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

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

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

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

Combine -4x and 2x to get -2x.

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

Add 4 and 1 to get 5.

$$2x^{2}-2x+5=x+7$$

Subtract x from both sides.

$$2x^{2}-2x+5-x=7$$

Combine -2x and -x to get -3x.

$$2x^{2}-3x+5=7$$

Subtract 5 from both sides.

$$2x^{2}-3x=7-5$$

Subtract 5 from 7 to get 2.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide 2 by 2.

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

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=1+\frac{9}{16}$$

Add 1 to \frac{9}{16}.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{25}{16}$$

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

$$\left(x-\frac{3}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{5}{4} x-\frac{3}{4}=-\frac{5}{4}$$

Add \frac{3}{4} to both sides of the equation.

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