Completing the Square when a = 1

A quadratic equation is an equation that contains a squared variable as its highest power on any variable. The general form of a quadratic equation is:

ax² + bx + c = 0

Where a, b, and c are constants and a ≠ 0. In other words there must be a x² term.

Some examples are:

x² + 3x - 3 = 0
4x² + 9 = 0 (Where b = 0)
x² + 5x = 0 (where c = 0)

One way to solve a quadratic equation is by completing the square.

ax² + bx + c = 0 → (x- r)² = S

Where r and s are constants.

This discussion will focus on completing the square when a, the x² -coefficient, is 1.

Let's solve the following equation by completing the square: x² = 6x - 7
Step 1: Write the equation in the general form ax² + bx + c = 0.	x² - 6x + 5 = 0
Step 2: Move c, the constant term, to the right-hand side of the equation.	c = 5 x² - 6x = -5
Step 3: Divide b, the x-coefficient, by two and square the result.	b = -6 $- \frac{6}{2} = - 3 \to r$ (-3)² = 9
Step 4: Add the result from Step 3 to both sides of the equation.	x² - 6x + 9 = -5 + 9
Step 5: Rewrite the left-hand side as a perfect square and simplify the right-hand side. When rewriting in perfect square format the value in the parentheses is the b, x-coefficient, divided by 2 as found in Step 3.	(x - 3)² = -5 + 9 (x - 3)² = 4
Now that the square has been completed, solve for x.
Step 6: Take the square root of both sides of the equation. Remember that when taking the square root on the right-hand side the answer can be positive or negative.	x - 3 = ± 2
Step 7: Solve for x.	x = 3 ± 2 x = 5 or x = 1

Example 1: x² + 2x - 5 = 0
Step 1: Write the equation in the general form ax² + bx + c = 0. This equation is already in the proper form.	x² + 2x - 5 = 0
Step 2: Move c, the constant term, to the right-hand side of the equation.	c = -5 x² + 2x = 5
Step 3: Divide b, the x -coefficient, by two and square the result.	b = 2 $\frac{2}{2} = 1 \to r$ (1)² = 1
Step 4: Add the result from Step 3 to both sides of the equation.	x² + 2x + 1 = 5 + 1
Step 5: Rewrite the left-hand side as a perfect square and simplify the right-hand side. When rewriting in perfect square format the value in the parentheses is the b, x-coefficient, divided by 2 as found in Step 3.	(x + 1)² = 5 + 1 (x + 1)² = 6
Now that the square has been completed, solve for x.
Step 6: Take the square root of both sides of the equation. Remember that when taking the square root on the right-hand side the answer can be positive or negative.	$x + 1 = \pm \sqrt{6}$
Step 7: Solve for x.	$x = - 1 \pm \sqrt{6}$

Example 2: $x^{2} = \frac{1}{2} x$
Step 1: Write the equation in the general form ax² + bx + c = 0.	$x^{2} - \frac{1}{2} x = 0$
Step 2: Move c, the constant term, to the right-hand side of the equation. There is no c in this equation.	$x^{2} - \frac{1}{2} x = 0$
Step 3: Divide b, the x-coefficient, by two and square the result.	$b = - \frac{1}{2}$ $- \frac{1}{2} \div 2 = - \frac{1}{4} \to r$ ${(- \frac{1}{4})}^{2} = \frac{1}{16}$
Step 4: Add the result from Step 3 to both sides of the equation.	$x^{2} - \frac{1}{2} x + \frac{1}{16} = 0 + \frac{1}{16}$
Step 5: Rewrite the left-hand side as a perfect square and simplify the right-hand side. When rewriting in perfect square format the value in the parentheses is the b, x-coefficient, divided by 2 as found in Step 3.	${(x - \frac{1}{4})}^{2} = 0 + \frac{1}{16}$ ${(x - \frac{1}{4})}^{2} = \frac{1}{16}$
Now that the square has been completed, solve for x.
Step 6: Take the square root of both sides of the equation. Remember that when taking the square root on the right-hand side the answer can be positive or negative.	$x - \frac{1}{4} = \pm \sqrt{\frac{1}{16}}$
Step 7: Solve for x.	$x = \frac{1}{4} \pm \frac{1}{4}$ $x = \frac{1}{2} o r x = 0$

PART II of Completing the Square addresses the case when a ≠ 1.

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