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

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

Solve for x

$$x = \frac{2 \sqrt{7} - 2}{3} \approx 1.097167541$$

$$x=\frac{-2\sqrt{7}-2}{3}\approx -2.430500874$$

Steps Using the Quadratic Formula

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

$$9+x^{2}=4x^{2}+4x+1$$

Subtract 4x^{2} from both sides.

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

Combine x^{2} and -4x^{2} to get -3x^{2}.

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

Subtract 4x from both sides.

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

Subtract 1 from both sides.

$$9-3x^{2}-4x-1=0$$

Subtract 1 from 9 to get 8.

$$8-3x^{2}-4x=0$$

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.

$$-3x^{2}-4x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -4 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-3\right)\times 8}}{2\left(-3\right)}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-3\right)\times 8}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-4\right)±\sqrt{16+12\times 8}}{2\left(-3\right)}$$

Multiply 12 times 8.

$$x=\frac{-\left(-4\right)±\sqrt{16+96}}{2\left(-3\right)}$$

Add 16 to 96.

$$x=\frac{-\left(-4\right)±\sqrt{112}}{2\left(-3\right)}$$

Take the square root of 112.

$$x=\frac{-\left(-4\right)±4\sqrt{7}}{2\left(-3\right)}$$

The opposite of -4 is 4.

$$x=\frac{4±4\sqrt{7}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{4±4\sqrt{7}}{-6}$$

Now solve the equation x=\frac{4±4\sqrt{7}}{-6} when ± is plus. Add 4 to 4\sqrt{7}.

$$x=\frac{4\sqrt{7}+4}{-6}$$

Divide 4+4\sqrt{7} by -6.

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

Now solve the equation x=\frac{4±4\sqrt{7}}{-6} when ± is minus. Subtract 4\sqrt{7} from 4.

$$x=\frac{4-4\sqrt{7}}{-6}$$

Divide 4-4\sqrt{7} by -6.

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

The equation is now solved.

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

Steps for Completing the Square

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

$$9+x^{2}=4x^{2}+4x+1$$

Subtract 4x^{2} from both sides.

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

Combine x^{2} and -4x^{2} to get -3x^{2}.

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

Subtract 4x from both sides.

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

Subtract 9 from both sides.

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

Subtract 9 from 1 to get -8.

$$-3x^{2}-4x=-8$$

Divide both sides by -3.

$$\frac{-3x^{2}-4x}{-3}=-\frac{8}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\left(-\frac{4}{-3}\right)x=-\frac{8}{-3}$$

Divide -4 by -3.

$$x^{2}+\frac{4}{3}x=-\frac{8}{-3}$$

Divide -8 by -3.

$$x^{2}+\frac{4}{3}x=\frac{8}{3}$$

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

$$x^{2}+\frac{4}{3}x+\left(\frac{2}{3}\right)^{2}=\frac{8}{3}+\left(\frac{2}{3}\right)^{2}$$

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

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{8}{3}+\frac{4}{9}$$

Add \frac{8}{3} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{28}{9}$$

Factor x^{2}+\frac{4}{3}x+\frac{4}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{2}{3}\right)^{2}=\frac{28}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{2}{3}\right)^{2}}=\sqrt{\frac{28}{9}}$$

Simplify.

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

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

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