Solve 9=9x^2+(x+1)^2

$$9 = 9 x ^ { 2 } + ( x + 1 ) ^ { 2 }$$

Solve for x

$$x=-1$$

$$x=\frac{4}{5}=0.8$$

Steps Using Factoring By Grouping

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

$$9=9x^{2}+x^{2}+2x+1$$

Combine 9x^{2} and x^{2} to get 10x^{2}.

$$9=10x^{2}+2x+1$$

Swap sides so that all variable terms are on the left hand side.

$$10x^{2}+2x+1=9$$

Subtract 9 from both sides.

$$10x^{2}+2x+1-9=0$$

Subtract 9 from 1 to get -8.

$$10x^{2}+2x-8=0$$

Divide both sides by 2.

$$5x^{2}+x-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=1 ab=5\left(-4\right)=-20$$

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 -20.

$$-1,20 -2,10 -4,5$$

Calculate the sum for each pair.

$$-1+20=19 -2+10=8 -4+5=1$$

The solution is the pair that gives sum 1.

$$a=-4 b=5$$

Rewrite 5x^{2}+x-4 as \left(5x^{2}-4x\right)+\left(5x-4\right).

$$\left(5x^{2}-4x\right)+\left(5x-4\right)$$

Factor out x in 5x^{2}-4x.

$$x\left(5x-4\right)+5x-4$$

Factor out common term 5x-4 by using distributive property.

$$\left(5x-4\right)\left(x+1\right)$$

To find equation solutions, solve 5x-4=0 and x+1=0.

$$x=\frac{4}{5} x=-1$$

Steps Using the Quadratic Formula

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

$$9=9x^{2}+x^{2}+2x+1$$

Combine 9x^{2} and x^{2} to get 10x^{2}.

$$9=10x^{2}+2x+1$$

Swap sides so that all variable terms are on the left hand side.

$$10x^{2}+2x+1=9$$

Subtract 9 from both sides.

$$10x^{2}+2x+1-9=0$$

Subtract 9 from 1 to get -8.

$$10x^{2}+2x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, 2 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\times 10\left(-8\right)}}{2\times 10}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 10\left(-8\right)}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-2±\sqrt{4-40\left(-8\right)}}{2\times 10}$$

Multiply -40 times -8.

$$x=\frac{-2±\sqrt{4+320}}{2\times 10}$$

Add 4 to 320.

$$x=\frac{-2±\sqrt{324}}{2\times 10}$$

Take the square root of 324.

$$x=\frac{-2±18}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{-2±18}{20}$$

Now solve the equation x=\frac{-2±18}{20} when ± is plus. Add -2 to 18.

$$x=\frac{16}{20}$$

Reduce the fraction \frac{16}{20} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{-2±18}{20} when ± is minus. Subtract 18 from -2.

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

Divide -20 by 20.

$$x=-1$$

The equation is now solved.

$$x=\frac{4}{5} x=-1$$

Steps for Completing the Square

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

$$9=9x^{2}+x^{2}+2x+1$$

Combine 9x^{2} and x^{2} to get 10x^{2}.

$$9=10x^{2}+2x+1$$

Swap sides so that all variable terms are on the left hand side.

$$10x^{2}+2x+1=9$$

Subtract 1 from both sides.

$$10x^{2}+2x=9-1$$

Subtract 1 from 9 to get 8.

$$10x^{2}+2x=8$$

Divide both sides by 10.

$$\frac{10x^{2}+2x}{10}=\frac{8}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$x^{2}+\frac{2}{10}x=\frac{8}{10}$$

Reduce the fraction \frac{2}{10} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{1}{5}x=\frac{8}{10}$$

Reduce the fraction \frac{8}{10} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{1}{5}x=\frac{4}{5}$$

Divide \frac{1}{5}, the coefficient of the x term, by 2 to get \frac{1}{10}. Then add the square of \frac{1}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{5}x+\left(\frac{1}{10}\right)^{2}=\frac{4}{5}+\left(\frac{1}{10}\right)^{2}$$

Square \frac{1}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{5}x+\frac{1}{100}=\frac{4}{5}+\frac{1}{100}$$

Add \frac{4}{5} to \frac{1}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{5}x+\frac{1}{100}=\frac{81}{100}$$

Factor x^{2}+\frac{1}{5}x+\frac{1}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{10}\right)^{2}=\frac{81}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{10}\right)^{2}}=\sqrt{\frac{81}{100}}$$

Simplify.

$$x+\frac{1}{10}=\frac{9}{10} x+\frac{1}{10}=-\frac{9}{10}$$

Subtract \frac{1}{10} from both sides of the equation.

$$x=\frac{4}{5} x=-1$$