Solve (3+x)^2+x^2=29

$$( 3 + x ) ^ { 2 } + x ^ { 2 } = 29$$

Solve for x

$$x=-5$$

$$x=2$$

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}.

$$9+6x+x^{2}+x^{2}=29$$

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

$$9+6x+2x^{2}=29$$

Subtract 29 from both sides.

$$9+6x+2x^{2}-29=0$$

Subtract 29 from 9 to get -20.

$$-20+6x+2x^{2}=0$$

Divide both sides by 2.

$$-10+3x+x^{2}=0$$

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

$$x^{2}+3x-10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-10. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\left(-10\right)=-10$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -10.

$$-1,10 -2,5$$

Calculate the sum for each pair.

$$-1+10=9 -2+5=3$$

The solution is the pair that gives sum 3.

$$a=-2 b=5$$

Rewrite x^{2}+3x-10 as \left(x^{2}-2x\right)+\left(5x-10\right).

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

Factor out x in the first and 5 in the second group.

$$x\left(x-2\right)+5\left(x-2\right)$$

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

$$\left(x-2\right)\left(x+5\right)$$

To find equation solutions, solve x-2=0 and x+5=0.

$$x=2 x=-5$$

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}.

$$9+6x+x^{2}+x^{2}=29$$

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

$$9+6x+2x^{2}=29$$

Subtract 29 from both sides.

$$9+6x+2x^{2}-29=0$$

Subtract 29 from 9 to get -20.

$$-20+6x+2x^{2}=0$$

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.

$$2x^{2}+6x-20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 6 for b, and -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 2\left(-20\right)}}{2\times 2}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 2\left(-20\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-6±\sqrt{36-8\left(-20\right)}}{2\times 2}$$

Multiply -8 times -20.

$$x=\frac{-6±\sqrt{36+160}}{2\times 2}$$

Add 36 to 160.

$$x=\frac{-6±\sqrt{196}}{2\times 2}$$

Take the square root of 196.

$$x=\frac{-6±14}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-6±14}{4}$$

Now solve the equation x=\frac{-6±14}{4} when ± is plus. Add -6 to 14.

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

Divide 8 by 4.

$$x=2$$

Now solve the equation x=\frac{-6±14}{4} when ± is minus. Subtract 14 from -6.

$$x=-\frac{20}{4}$$

Divide -20 by 4.

$$x=-5$$

The equation is now solved.

$$x=2 x=-5$$

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}.

$$9+6x+x^{2}+x^{2}=29$$

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

$$9+6x+2x^{2}=29$$

Subtract 9 from both sides.

$$6x+2x^{2}=29-9$$

Subtract 9 from 29 to get 20.

$$6x+2x^{2}=20$$

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.

$$2x^{2}+6x=20$$

Divide both sides by 2.

$$\frac{2x^{2}+6x}{2}=\frac{20}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{6}{2}x=\frac{20}{2}$$

Divide 6 by 2.

$$x^{2}+3x=\frac{20}{2}$$

Divide 20 by 2.

$$x^{2}+3x=10$$

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.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=10+\left(\frac{3}{2}\right)^{2}$$

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

$$x^{2}+3x+\frac{9}{4}=10+\frac{9}{4}$$

Add 10 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{49}{4}$$

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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{7}{2} x+\frac{3}{2}=-\frac{7}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=2 x=-5$$