Solve x_0^2-2x_{0}=3

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

Solve for x_0

$$x_{0}=-1$$

$$x_{0}=3$$

Steps Using Factoring

Subtract 3 from both sides.

$$x_{0}^{2}-2x_{0}-3=0$$

To solve the equation, factor x_{0}^{2}-2x_{0}-3 using formula x_{0}^{2}+\left(a+b\right)x_{0}+ab=\left(x_{0}+a\right)\left(x_{0}+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 negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=-3 b=1$$

Rewrite factored expression \left(x_{0}+a\right)\left(x_{0}+b\right) using the obtained values.

$$\left(x_{0}-3\right)\left(x_{0}+1\right)$$

To find equation solutions, solve x_{0}-3=0 and x_{0}+1=0.

$$x_{0}=3 x_{0}=-1$$

Steps Using Factoring By Grouping

Subtract 3 from both sides.

$$x_{0}^{2}-2x_{0}-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x_{0}^{2}+ax_{0}+bx_{0}-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 negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=-3 b=1$$

Rewrite x_{0}^{2}-2x_{0}-3 as \left(x_{0}^{2}-3x_{0}\right)+\left(x_{0}-3\right).

$$\left(x_{0}^{2}-3x_{0}\right)+\left(x_{0}-3\right)$$

Factor out x_{0} in x_{0}^{2}-3x_{0}.

$$x_{0}\left(x_{0}-3\right)+x_{0}-3$$

Factor out common term x_{0}-3 by using distributive property.

$$\left(x_{0}-3\right)\left(x_{0}+1\right)$$

To find equation solutions, solve x_{0}-3=0 and x_{0}+1=0.

$$x_{0}=3 x_{0}=-1$$

Steps Using the Quadratic Formula

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.

$$x_{0}^{2}-2x_{0}=3$$

Subtract 3 from both sides of the equation.

$$x_{0}^{2}-2x_{0}-3=3-3$$

Subtracting 3 from itself leaves 0.

$$x_{0}^{2}-2x_{0}-3=0$$

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}.

$$x_{0}=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-3\right)}}{2}$$

Square -2.

$$x_{0}=\frac{-\left(-2\right)±\sqrt{4-4\left(-3\right)}}{2}$$

Multiply -4 times -3.

$$x_{0}=\frac{-\left(-2\right)±\sqrt{4+12}}{2}$$

Add 4 to 12.

$$x_{0}=\frac{-\left(-2\right)±\sqrt{16}}{2}$$

Take the square root of 16.

$$x_{0}=\frac{-\left(-2\right)±4}{2}$$

The opposite of -2 is 2.

$$x_{0}=\frac{2±4}{2}$$

Now solve the equation x_{0}=\frac{2±4}{2} when ± is plus. Add 2 to 4.

$$x_{0}=\frac{6}{2}$$

Divide 6 by 2.

$$x_{0}=3$$

Now solve the equation x_{0}=\frac{2±4}{2} when ± is minus. Subtract 4 from 2.

$$x_{0}=-\frac{2}{2}$$

Divide -2 by 2.

$$x_{0}=-1$$

The equation is now solved.

$$x_{0}=3 x_{0}=-1$$

Steps for Completing the Square

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.

$$x_{0}^{2}-2x_{0}=3$$

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.

$$x_{0}^{2}-2x_{0}+1=3+1$$

Add 3 to 1.

$$x_{0}^{2}-2x_{0}+1=4$$

Factor x_{0}^{2}-2x_{0}+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}.

$$\left(x_{0}-1\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x_{0}-1\right)^{2}}=\sqrt{4}$$

Simplify.

$$x_{0}-1=2 x_{0}-1=-2$$

Add 1 to both sides of the equation.

$$x_{0}=3 x_{0}=-1$$