Solve 8x*(1+2x)=-1

$$8 x \cdot ( 1 + 2 x ) = - 1$$

Solve for x

$$x=-\frac{1}{4}=-0.25$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 8x by 1+2x.

$$8x+16x^{2}=-1$$

Add 1 to both sides.

$$8x+16x^{2}+1=0$$

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.

$$16x^{2}+8x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, 8 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 16}}{2\times 16}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 16}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-8±\sqrt{64-64}}{2\times 16}$$

Add 64 to -64.

$$x=\frac{-8±\sqrt{0}}{2\times 16}$$

Take the square root of 0.

$$x=-\frac{8}{2\times 16}$$

Multiply 2 times 16.

$$x=-\frac{8}{32}$$

Reduce the fraction \frac{-8}{32} to lowest terms by extracting and canceling out 8.

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

Steps for Completing the Square

Use the distributive property to multiply 8x by 1+2x.

$$8x+16x^{2}=-1$$

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.

$$16x^{2}+8x=-1$$

Divide both sides by 16.

$$\frac{16x^{2}+8x}{16}=-\frac{1}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$x^{2}+\frac{8}{16}x=-\frac{1}{16}$$

Reduce the fraction \frac{8}{16} to lowest terms by extracting and canceling out 8.

$$x^{2}+\frac{1}{2}x=-\frac{1}{16}$$

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=-\frac{1}{16}+\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{-1+1}{16}$$

Add -\frac{1}{16} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=0$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{4}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x+\frac{1}{4}=0 x+\frac{1}{4}=0$$

Subtract \frac{1}{4} from both sides of the equation.

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

The equation is now solved. Solutions are the same.

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