Solve x^2+2x-3/x-1=0

$$\frac { x ^ { 2 } + 2 x - 3 } { x - 1 } = 0$$

Solve for x

$$x=-3$$

Steps Using Factoring

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by x-1.

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

To solve the equation, factor x^{2}+2x-3 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=2 ab=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=3$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-1\right)\left(x+3\right)$$

To find equation solutions, solve x-1=0 and x+3=0.

$$x=1 x=-3$$

Variable x cannot be equal to 1.

$$x=-3$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by x-1.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-3\right)=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=3$$

Rewrite x^{2}+2x-3 as \left(x^{2}-x\right)+\left(3x-3\right).

$$\left(x^{2}-x\right)+\left(3x-3\right)$$

Factor out x in the first and 3 in the second group.

$$x\left(x-1\right)+3\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(x+3\right)$$

To find equation solutions, solve x-1=0 and x+3=0.

$$x=1 x=-3$$

Variable x cannot be equal to 1.

$$x=-3$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by x-1.

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

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

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

Square 2.

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

Multiply -4 times -3.

$$x=\frac{-2±\sqrt{4+12}}{2}$$

Add 4 to 12.

$$x=\frac{-2±\sqrt{16}}{2}$$

Take the square root of 16.

$$x=\frac{-2±4}{2}$$

Now solve the equation x=\frac{-2±4}{2} when ± is plus. Add -2 to 4.

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

Divide 2 by 2.

$$x=1$$

Now solve the equation x=\frac{-2±4}{2} when ± is minus. Subtract 4 from -2.

$$x=-\frac{6}{2}$$

Divide -6 by 2.

$$x=-3$$

The equation is now solved.

$$x=1 x=-3$$

Variable x cannot be equal to 1.

$$x=-3$$

Steps for Completing the Square

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by x-1.

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

Add 3 to both sides. Anything plus zero gives itself.

$$x^{2}+2x=3$$

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

$$x^{2}+2x+1^{2}=3+1^{2}$$

Square 1.

$$x^{2}+2x+1=3+1$$

Add 3 to 1.

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

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

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+1=2 x+1=-2$$

Subtract 1 from both sides of the equation.

$$x=1 x=-3$$

Variable x cannot be equal to 1.

$$x=-3$$