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

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

Solve for x

$$x = \frac{\sqrt{109} + 1}{6} \approx 1.906717751$$

$$x=\frac{1-\sqrt{109}}{6}\approx -1.573384418$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x by 1-2x.

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

Subtract 9 from both sides.

$$x^{2}-9=x-2x^{2}$$

Subtract x from both sides.

$$x^{2}-9-x=-2x^{2}$$

Add 2x^{2} to both sides.

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

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

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -1 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 3\left(-9\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-1\right)±\sqrt{1-12\left(-9\right)}}{2\times 3}$$

Multiply -12 times -9.

$$x=\frac{-\left(-1\right)±\sqrt{1+108}}{2\times 3}$$

Add 1 to 108.

$$x=\frac{-\left(-1\right)±\sqrt{109}}{2\times 3}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{109}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{1±\sqrt{109}}{6}$$

Now solve the equation x=\frac{1±\sqrt{109}}{6} when ± is plus. Add 1 to \sqrt{109}.

$$x=\frac{\sqrt{109}+1}{6}$$

Now solve the equation x=\frac{1±\sqrt{109}}{6} when ± is minus. Subtract \sqrt{109} from 1.

$$x=\frac{1-\sqrt{109}}{6}$$

The equation is now solved.

$$x=\frac{\sqrt{109}+1}{6} x=\frac{1-\sqrt{109}}{6}$$

Steps for Completing the Square

Use the distributive property to multiply x by 1-2x.

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

Subtract x from both sides.

$$x^{2}-x=9-2x^{2}$$

Add 2x^{2} to both sides.

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

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

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

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide 9 by 3.

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

Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=3+\left(-\frac{1}{6}\right)^{2}$$

Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=3+\frac{1}{36}$$

Add 3 to \frac{1}{36}.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{109}{36}$$

Factor x^{2}-\frac{1}{3}x+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{6}\right)^{2}=\frac{109}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{109}{36}}$$

Simplify.

$$x-\frac{1}{6}=\frac{\sqrt{109}}{6} x-\frac{1}{6}=-\frac{\sqrt{109}}{6}$$

Add \frac{1}{6} to both sides of the equation.

$$x=\frac{\sqrt{109}+1}{6} x=\frac{1-\sqrt{109}}{6}$$