Solve 8x=1-x^2

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

Solve for x (complex solution)

$$x=\sqrt{17}-4\approx 0.123105626$$

$$x=-\left(\sqrt{17}+4\right)\approx -8.123105626$$

Steps Using the Quadratic Formula

Subtract 1 from both sides.

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

Add x^{2} to both sides.

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

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.

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

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

$$x=\frac{-8±\sqrt{8^{2}-4\left(-1\right)}}{2}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-1\right)}}{2}$$

Multiply -4 times -1.

$$x=\frac{-8±\sqrt{64+4}}{2}$$

Add 64 to 4.

$$x=\frac{-8±\sqrt{68}}{2}$$

Take the square root of 68.

$$x=\frac{-8±2\sqrt{17}}{2}$$

Now solve the equation x=\frac{-8±2\sqrt{17}}{2} when ± is plus. Add -8 to 2\sqrt{17}.

$$x=\frac{2\sqrt{17}-8}{2}$$

Divide -8+2\sqrt{17} by 2.

$$x=\sqrt{17}-4$$

Now solve the equation x=\frac{-8±2\sqrt{17}}{2} when ± is minus. Subtract 2\sqrt{17} from -8.

$$x=\frac{-2\sqrt{17}-8}{2}$$

Divide -8-2\sqrt{17} by 2.

$$x=-\sqrt{17}-4$$

The equation is now solved.

$$x=\sqrt{17}-4 x=-\sqrt{17}-4$$

Steps for Completing the Square

Add x^{2} to both sides.

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

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.

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

Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+8x+4^{2}=1+4^{2}$$

Square 4.

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

Add 1 to 16.

$$x^{2}+8x+16=17$$

Factor x^{2}+8x+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}.

$$\left(x+4\right)^{2}=17$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+4\right)^{2}}=\sqrt{17}$$

Simplify.

$$x+4=\sqrt{17} x+4=-\sqrt{17}$$

Subtract 4 from both sides of the equation.

$$x=\sqrt{17}-4 x=-\sqrt{17}-4$$

Solve for x

$$x=\sqrt{17}-4\approx 0.123105626$$

$$x=-\sqrt{17}-4\approx -8.123105626$$

Steps Using the Quadratic Formula

Subtract 1 from both sides.

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

Add x^{2} to both sides.

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

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.

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

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

$$x=\frac{-8±\sqrt{8^{2}-4\left(-1\right)}}{2}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-1\right)}}{2}$$

Multiply -4 times -1.

$$x=\frac{-8±\sqrt{64+4}}{2}$$

Add 64 to 4.

$$x=\frac{-8±\sqrt{68}}{2}$$

Take the square root of 68.

$$x=\frac{-8±2\sqrt{17}}{2}$$

Now solve the equation x=\frac{-8±2\sqrt{17}}{2} when ± is plus. Add -8 to 2\sqrt{17}.

$$x=\frac{2\sqrt{17}-8}{2}$$

Divide -8+2\sqrt{17} by 2.

$$x=\sqrt{17}-4$$

Now solve the equation x=\frac{-8±2\sqrt{17}}{2} when ± is minus. Subtract 2\sqrt{17} from -8.

$$x=\frac{-2\sqrt{17}-8}{2}$$

Divide -8-2\sqrt{17} by 2.

$$x=-\sqrt{17}-4$$

The equation is now solved.

$$x=\sqrt{17}-4 x=-\sqrt{17}-4$$

Steps for Completing the Square

Add x^{2} to both sides.

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

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.

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

Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+8x+4^{2}=1+4^{2}$$

Square 4.

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

Add 1 to 16.

$$x^{2}+8x+16=17$$

Factor x^{2}+8x+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}.

$$\left(x+4\right)^{2}=17$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+4\right)^{2}}=\sqrt{17}$$

Simplify.

$$x+4=\sqrt{17} x+4=-\sqrt{17}$$

Subtract 4 from both sides of the equation.

$$x=\sqrt{17}-4 x=-\sqrt{17}-4$$