Solve 9=(2x+8x+3)^4/2

$$9 = ( 2 x + 8 x + 3 ) ^ { 4 / 2 }$$

Solve for x

$$x=0$$

$$x=-\frac{3}{5}=-0.6$$

Steps Using Factoring

Combine 2x and 8x to get 10x.

$$9=\left(10x+3\right)^{\frac{4}{2}}$$

Divide 4 by 2 to get 2.

$$9=\left(10x+3\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(10x+3\right)^{2}.

$$9=100x^{2}+60x+9$$

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

$$100x^{2}+60x+9=9$$

Subtract 9 from both sides.

$$100x^{2}+60x+9-9=0$$

Subtract 9 from 9 to get 0.

$$100x^{2}+60x=0$$

Factor out x.

$$x\left(100x+60\right)=0$$

To find equation solutions, solve x=0 and 100x+60=0.

$$x=0 x=-\frac{3}{5}$$

Steps Using the Quadratic Formula

Combine 2x and 8x to get 10x.

$$9=\left(10x+3\right)^{\frac{4}{2}}$$

Divide 4 by 2 to get 2.

$$9=\left(10x+3\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(10x+3\right)^{2}.

$$9=100x^{2}+60x+9$$

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

$$100x^{2}+60x+9=9$$

Subtract 9 from both sides.

$$100x^{2}+60x+9-9=0$$

Subtract 9 from 9 to get 0.

$$100x^{2}+60x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 100 for a, 60 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-60±\sqrt{60^{2}}}{2\times 100}$$

Take the square root of 60^{2}.

$$x=\frac{-60±60}{2\times 100}$$

Multiply 2 times 100.

$$x=\frac{-60±60}{200}$$

Now solve the equation x=\frac{-60±60}{200} when ± is plus. Add -60 to 60.

$$x=\frac{0}{200}$$

Divide 0 by 200.

$$x=0$$

Now solve the equation x=\frac{-60±60}{200} when ± is minus. Subtract 60 from -60.

$$x=-\frac{120}{200}$$

Reduce the fraction \frac{-120}{200} to lowest terms by extracting and canceling out 40.

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

The equation is now solved.

$$x=0 x=-\frac{3}{5}$$

Steps for Completing the Square

Combine 2x and 8x to get 10x.

$$9=\left(10x+3\right)^{\frac{4}{2}}$$

Divide 4 by 2 to get 2.

$$9=\left(10x+3\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(10x+3\right)^{2}.

$$9=100x^{2}+60x+9$$

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

$$100x^{2}+60x+9=9$$

Subtract 9 from both sides.

$$100x^{2}+60x=9-9$$

Subtract 9 from 9 to get 0.

$$100x^{2}+60x=0$$

Divide both sides by 100.

$$\frac{100x^{2}+60x}{100}=\frac{0}{100}$$

Dividing by 100 undoes the multiplication by 100.

$$x^{2}+\frac{60}{100}x=\frac{0}{100}$$

Reduce the fraction \frac{60}{100} to lowest terms by extracting and canceling out 20.

$$x^{2}+\frac{3}{5}x=\frac{0}{100}$$

Divide 0 by 100.

$$x^{2}+\frac{3}{5}x=0$$

Divide \frac{3}{5}, the coefficient of the x term, by 2 to get \frac{3}{10}. Then add the square of \frac{3}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{5}x+\left(\frac{3}{10}\right)^{2}=\left(\frac{3}{10}\right)^{2}$$

Square \frac{3}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{5}x+\frac{9}{100}=\frac{9}{100}$$

Factor x^{2}+\frac{3}{5}x+\frac{9}{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}.

$$\left(x+\frac{3}{10}\right)^{2}=\frac{9}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{10}\right)^{2}}=\sqrt{\frac{9}{100}}$$

Simplify.

$$x+\frac{3}{10}=\frac{3}{10} x+\frac{3}{10}=-\frac{3}{10}$$

Subtract \frac{3}{10} from both sides of the equation.

$$x=0 x=-\frac{3}{5}$$