# Solving Quadratic Equations II: Taking Square Roots

A quadratic equation can be solved by taking the square root of both sides of the equation. This method uses the square root property,
$y^2=z \to y=\pm \sqrt{z}$
Before taking the square root, the equation must be arranged with the x2 term isolated on the left- hand side of the equation and its coefficient reduced to 1. There are four steps in solving quadratic equations by this method:

Step 1: Isolate the $x^2$ and $x$ terms. Use the addition and subtraction and isolate the $x^2$ and $x$ terms on the left-hand side of the equation. Then, use the multiplication and division axioms to eliminate the coefficient from the $x^2$ term.

Step 2: Make the coefficient on the $x^2$ term equal to $1$. Use multiplication or division to eliminate the coefficient from the $x^2$ term.

Step 3: Complete the square. To complete the square, take the coefficient of the $x$ term, square it, and divide it by 4.

Step 4: Solve the equation in step 3 by taking the square root of both sides of the equation.

Example 1: $x^2+4x-7=0$

Step 1: Isolate the $x^2$ and $x$ terms.
$x^2+4x=7$

Step 2: Make the coefficient on the $x^2$ term equal to $1$.\\
It is already $1$.

Step 3: Complete the square.
$x^2+4x+4=+4+7=11$
$\to (x+2)^2=11$
Step 4: Solve the equation in step 3 by taking the square root of both sides of the equation.
$(x+2)^2=11 \to \left\{ \begin{array}{l} x+2=+\sqrt{11} \to x=-2+\sqrt{11} \\ x+2=-\sqrt{11} \to x=-2-\sqrt{11} \end{array} \right.$

Example 2: $x^2-3x+2=0$

Step 1: Isolate the $x^2$ and $x$ terms.
$x^2-3x=-2$

Step 2: Make the coefficient on the $x^2$ term equal to $1$.\\
It is already $1$.

Step 3: Complete the square.
$x^2-3x+\frac{9}{4}=+\frac{9}{4}-2=\frac{1}{4}$
$\to \left( x-\frac{3}{2} \right)^2=\frac{1}{4}$
Step 4: Solve the equation in step 3 by taking the square root of both sides of the equation.
$\left( x-\frac{3}{2} \right)^2=\frac{1}{4} \to \left\{ \begin{array}{l} x-\frac{3}{2}=+\frac{1}{2} \to x=2 \\ x-\frac{3}{2}=-\frac{1}{4} \to x=1 \end{array} \right.$

Example 3: $-2x^2+2x+12=0$

Step 1: Isolate the $x^2$ and $x$ terms.
$-2x^2+2x=-12$

Step 2: Make the coefficient on the $x^2$ term equal to $1$.\\
$x^2-x=6$

Step 3: Complete the square.
$x^2-x+\frac{1}{4}=+\frac{1}{4}+6=\frac{25}{4}$
$\to \left( x-\frac{1}{2} \right)^2=\frac{25}{4}$
Step 4: Solve the equation in step 3 by taking the square root of both sides of the equation.
$\left( x-\frac{1}{2} \right)^2=\frac{25}{4} \to \left\{ \begin{array}{l} x-\frac{1}{2}=+\frac{5}{2} \to x=3 \\ x-\frac{1}{2}=-\frac{5}{2} \to x=-2 \end{array} \right.$

Example 4: $15-3x^2=4x$

Step 1: Isolate the $x^2$ and $x$ terms.
$-3x^2-4x=-15$

Step 2: Make the coefficient on the $x^2$ term equal to $1$.\\
$x^2+\frac{4}{3}x=5$

Step 3: Complete the square.
$x^2+\frac{4}{3}x+\frac{4}{9}=+\frac{4}{9}+5=\frac{49}{9}$
$\to \left( x+\frac{2}{3} \right)^2=\frac{49}{9}$
Step 4: Solve the equation in step 3 by taking the square root of both sides of the equation.

$\left( x+\frac{2}{3} \right)^2=\frac{49}{9} \to \left\{ \begin{array}{l} x+\frac{2}{3}=+\frac{7}{3} \to x=\frac{5}{3} \\ x+\frac{2}{3}=-\frac{7}{3} \to x=-3 \end{array} \right.$

## 2 thoughts on “Solving Quadratic Equations II: Taking Square Roots”

1. Rob says:

I'm confused about step 3 in example 1. You've added 16 to both sides, which I assume you got from squaring the coefficient of the x term, 4. But shouldn't you then divide 16 by 4, so that you add 4 to both sides? I don't understand how (x + 4)^2 is the same as x^2 + 4x + 16.

1. Apar says:

Hi Rob,
I apologize, my solution was wrong. I also changed the problem from $x^2+4x+7=0$ to $x^2+4x-7=0$ (just to have real answers)