# 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. $x+1=±\sqrt{6}$ Step 7: Solve for x. $x=-1±\sqrt{6}$

 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. ${b}=-\frac{1}{2}$ $-\frac{1}{2}÷2={-}\frac{{1}}{{4}}\to {r}$ ${\left(-\frac{1}{4}\right)}^{2}=\frac{{1}}{{16}}$ 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. ${\left(x{-}\frac{{1}}{{4}}\right)}^{2}=0+\frac{1}{16}$ ${\left(x{-}\frac{{1}}{{4}}\right)}^{2}=\frac{{1}}{{16}}$ 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. $x-\frac{1}{4}=±\sqrt{\frac{1}{16}}$ Step 7: Solve for x. $x=\frac{1}{4}±\frac{1}{4}$

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

 Related Links: Math algebra Completing the Square when a ≠ 1 Factoring Quadratic Equations when a equals 1 Algebra Topics

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