Solve x^2+40x/x^2+3x-2=0

$$\frac { x ^ { 2 } + 40 x } { x ^ { 2 } + 3 x - 2 } = 0$$

Solve for x

$$x=-40$$

$$x=0$$

Steps Using Factoring

Multiply both sides of the equation by \left(x-\left(-\frac{1}{2}\sqrt{17}-\frac{3}{2}\right)\right)\left(x-\left(\frac{1}{2}\sqrt{17}-\frac{3}{2}\right)\right).

$$x^{2}+40x=0$$

Factor out x.

$$x\left(x+40\right)=0$$

To find equation solutions, solve x=0 and x+40=0.

$$x=0 x=-40$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by \left(x-\left(-\frac{1}{2}\sqrt{17}-\frac{3}{2}\right)\right)\left(x-\left(\frac{1}{2}\sqrt{17}-\frac{3}{2}\right)\right).

$$x^{2}+40x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 40 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-40±\sqrt{40^{2}}}{2}$$

Take the square root of 40^{2}.

$$x=\frac{-40±40}{2}$$

Now solve the equation x=\frac{-40±40}{2} when ± is plus. Add -40 to 40.

$$x=\frac{0}{2}$$

Divide 0 by 2.

$$x=0$$

Now solve the equation x=\frac{-40±40}{2} when ± is minus. Subtract 40 from -40.

$$x=-\frac{80}{2}$$

Divide -80 by 2.

$$x=-40$$

The equation is now solved.

$$x=0 x=-40$$

Steps for Completing the Square

Multiply both sides of the equation by \left(x-\left(-\frac{1}{2}\sqrt{17}-\frac{3}{2}\right)\right)\left(x-\left(\frac{1}{2}\sqrt{17}-\frac{3}{2}\right)\right).

$$x^{2}+40x=0$$

Divide 40, the coefficient of the x term, by 2 to get 20. Then add the square of 20 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+40x+20^{2}=20^{2}$$

Square 20.

$$x^{2}+40x+400=400$$

Factor x^{2}+40x+400. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+20\right)^{2}=400$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+20\right)^{2}}=\sqrt{400}$$

Simplify.

$$x+20=20 x+20=-20$$

Subtract 20 from both sides of the equation.

$$x=0 x=-40$$