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8. Quadratic Equations
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Algebra 1

8.03 Solve quadratics using square roots

Lesson
Practice

The square root property

We can solve quadratic equations in the form a(x-h)^2=k by isolating the perfect square, then taking the square root of both sides of the equation.

1\displaystyle a\left(x-h\right)^2\displaystyle =\displaystyle kGiven equation
2\displaystyle \left(x-h\right)^2\displaystyle =\displaystyle \frac{k}{a}Divide by a
3\displaystyle x-h\displaystyle =\displaystyle \pm\sqrt{\frac{k}{a}}Square root property
4\displaystyle x\displaystyle =\displaystyle h\pm\sqrt{\frac{k}{a}}Add h

Following these steps, we can see that if \dfrac{k}{a} is not negative, then the equation will have real solutions. Otherwise, the equation will have no real solutions.

Another thing to notice is that taking the square root of both sides introduces the \pm symbol. This is because we have a positive and a negative root. These come from the fact that x^2=(-1)^2\cdot x^2=(-x)^2 so \sqrt{x^2}=\pm x.

When the radicand is not a perfect square, we need to simplify the expression. Radical expressions are written in simplified radical form if the radicand cannot be factored any further.

We can use the following facts to simplify radical expressions, for a, b \geq 0:

\displaystyle \sqrt{ab}\displaystyle =\displaystyle \sqrt{a} \sqrt{b} Multiplication property of radicals
\displaystyle \sqrt{\dfrac{a}{b}}\displaystyle =\displaystyle \dfrac{\sqrt{a}}{\sqrt{b}}Division property of radicals

We can simplify using properties of exponents, properties of radicals, or a perfect square factor:

Properties of exponents

A figure showing how to simplify square root of 24. First line: square root of 24 equals 24 raised to one half; Second line: equals left parenthesis 2 times 2 times 2 times 3 right parenthesis raised to one half; Third line: equals left parenthesis 2 squared times 2 times 3 right parenthesis raised to one half; Fourth line: equals left parenthesis 2 squared right parenthesis raised to one half times 2 raised to one half times 3 raised to one half; Fifth line: equals 2 times left parenthesis 2 times 3 right parenthesis raised to one half; Sixth line: equals 2 times 6 raised to one half; Seventh line: equals 2 times square root of 6.

When we simplified radicals using rational exponents we did the following. First, we converted the radical to a rational exponent. Then, we found the prime factors of 24 and applied properties of exponents to simplify.

Properties of radicals

A figure showing how to simplify square root of 24. First line: square root of 24 equals square root of 2 times 2 times 2 times 3; Second line: equals square root of 2 squared times 2 times 3; Third line: equals square root of 2 squared times square root of 2 times square root of 3; Fourth line: equals 2 times square root of 2 times 3; Fifth line: equals 2 times square root of 6.

We can follow a similar process for this method, except we can leave the expression in radical form. First, we find the prime factors of 24, then we can use properties of radicals to simplify.

Perfect square method

A figure showing how to simplify square root of 24. First line: square root of 24 equals square root of 4 times 6; Second line: equals square root of 4 times square root of 6; Third line: equals 2 times square root of 6.

This is the quickest method for simplifying a radical. Instead of finding all the prime factors of 24, we want to find the largest perfect square factor of 24. Then, we can use the multiplication property of radicals to simplify.

Examples

Example 1

Solve the following equations by using square roots:

a

x^2=9

Worked Solution
Create a strategy

In this equation we have 9 being equal to the square of x. This is equivalent to x being equal to the square root of 9.

Apply the idea

x=\pm 3

Reflect and check

Checking our answers:

(-3)^2=9

(3)^2=9

Both answers satisfy the equation.

b

4x^2-27=0

Worked Solution
Create a strategy

We can begin by isolating the variable, but we will need to simplify the radical since 27 is not a perfect square.

Apply the idea
\displaystyle 4x^2-27\displaystyle =\displaystyle 0Given equation
\displaystyle 4x^2\displaystyle =\displaystyle 27Addition property of equality
\displaystyle x^2\displaystyle =\displaystyle \dfrac{27}{4}Division property of equality
\displaystyle x\displaystyle =\displaystyle \pm\sqrt{\dfrac{27}{4}}Square root property
\displaystyle x\displaystyle =\displaystyle \pm \dfrac{\sqrt{27}}{\sqrt{4}}Divsion property of radicals

From here, we can factor 27. Our goal is to separate it into factors where one is a perfect square. 27=9\cdot 3 where 9 is a perfect square.

\displaystyle x\displaystyle =\displaystyle \pm \dfrac{\sqrt{9}\sqrt{3}}{\sqrt{4}}Multiplication property of radicals
\displaystyle x\displaystyle =\displaystyle \pm \dfrac{3\sqrt{3}}{2}Evaluate the radicals
Reflect and check

For nearly all of our work with solutions to functions and equations, it is standard practice to leave our final expression in exact form.

In questions involving applications of quadratics, we may be asked to evaluate the square root at the very end using a calculator, then approximate to a specific number of decimal places.

c

\left(x-2\right)^2-100=0

Worked Solution
Create a strategy

In order to use square roots to solve, the squared expression must be isolated. In this example we want to isolate the term \left(x-2\right)^2.

Apply the idea
\displaystyle \left(x-2\right)^2-100\displaystyle =\displaystyle 0Given equation
\displaystyle \left(x-2\right)^2\displaystyle =\displaystyle 100Add 100 to both sides
\displaystyle x-2\displaystyle =\displaystyle \pm 10Take the square root of both sides

This leaves us with two equations x-2=10 and x-2=-10. Add 2 to solve both equations and we find that the solutions are x=-8,\, x=12.

Reflect and check

Looking at the structure of this equation, we see it is in the form a(x-h)^2+k=0 which is the vertex form of its equivalent function. The vertex of this related parabola is at (2,-100), and we just found the zeros at (-8,0) and (12,0).

-10
-5
5
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-100
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d

\left(3x-8\right)^2=25

Worked Solution
Create a strategy

Since the squared expression is already isolated, we are ready to solve by taking square roots.

Apply the idea
\displaystyle \left(3x-8\right)^2\displaystyle =\displaystyle 25Given equation
\displaystyle 3x-8\displaystyle =\displaystyle \pm 5Square root
\displaystyle 3x\displaystyle =\displaystyle 8\pm 5Add 8 to both sides
\displaystyle x\displaystyle =\displaystyle \frac{8\pm 5}{3}Divide both sides by 3

This leaves us with two equations:

\displaystyle x=\dfrac{8-5}{3}\text{ and }x\displaystyle =\displaystyle \dfrac{8+5}{3}
\displaystyle x=\dfrac{3}{3}\text{ and }x\displaystyle =\displaystyle \dfrac{13}{3}

The solutions are x=1,\, x=\dfrac{13}{3}.

Reflect and check

We can check these answers by substituting them back into the original equation.

\displaystyle (3x-8)^2\displaystyle =\displaystyle 25Original equation
\displaystyle (3(1)-8)^2\displaystyle =\displaystyle 25Substitute x=1
\displaystyle (3-8)^2\displaystyle =\displaystyle 25Evaluate the multiplication
\displaystyle (-5)^2\displaystyle =\displaystyle 25Evaluate the subtraction
\displaystyle 25\displaystyle =\displaystyle 25Evaluate the exponent

Now checking the next solution:

\displaystyle (3x-8)^2\displaystyle =\displaystyle 25Original equation
\displaystyle \left(3\left(\dfrac{13}{3}\right)-8\right)^2\displaystyle =\displaystyle 25Substitute x=\frac{13}{3}
\displaystyle (13-8)^2\displaystyle =\displaystyle 25Evaluate the multiplication
\displaystyle (5)^2\displaystyle =\displaystyle 25Evaluate the subtraction
\displaystyle 25\displaystyle =\displaystyle 25Evaluate the exponent

Both answers satisfy the original equation.

Example 2

State a quadratic equation that has the given solutions.

a

x=-1\pm\sqrt{7}

Worked Solution
Create a strategy

We can work backwards from solving to find the equation that had the given solutions.

Apply the idea
\displaystyle x\displaystyle =\displaystyle -1\pm\sqrt{7}Given solutions
\displaystyle x+1\displaystyle =\displaystyle \pm\sqrt{7}Add 1 to both sides
\displaystyle (x+1)^2\displaystyle =\displaystyle 7Square both sides
Reflect and check

This is one equation, but we could also subtract 7 from both sides to get an equivalent equation with the same solutions.

(x+1)^2-7=0

b

x=\dfrac{5\pm\sqrt{10}}{3}

Worked Solution
Create a strategy

We can use a similar process as the previous problem, but this time we need to multiply both sides by 3 first.

Apply the idea
\displaystyle x\displaystyle =\displaystyle \dfrac{5\pm \sqrt{10}}{3}Given solutions
\displaystyle 3x\displaystyle =\displaystyle 5\pm \sqrt{10}Multiply 3 to both sides
\displaystyle 3x-5\displaystyle =\displaystyle \pm \sqrt{10}Subtract 5 from both sides
\displaystyle (3x-5)^2\displaystyle =\displaystyle 10Square both sides
Reflect and check

When completing the square, fractional solutions come from equations where a\neq 1. The 3 in the denominator came from the coefficient of x inside the parentheses.

Example 3

A square field has perpendicular lines drawn across it dividing it into 36 equal sized smaller squares. If the total area of the field is 225 square feet, determine the side length of one of the smaller squares.

Worked Solution
Create a strategy

We know that there are 36 smaller squares in total on a larger square grid, so there must be 6 by 6 smaller squares on the grid. If we let the side of a smaller square be x, then the side of the larger square can be 6x. This gives us the quadratic equation \left(6x\right)^2=225. We can then solve this equation by taking square roots.

Apply the idea
\displaystyle (6x)^2\displaystyle =\displaystyle 225Written equation
\displaystyle 6x\displaystyle =\displaystyle \pm 15Take the square root of both sides
\displaystyle x\displaystyle =\displaystyle \frac{\pm 15}{6}Divide by 6

Simplifying the expression gives us the solutions x=\pm 2.5 . We can exclude the negative solution as the length of the square must be positive. So the smaller square has a side length of 2.5 feet.

Reflect and check

In most real life applications, we will exclude the negative solution as it will be non-viable for the context.

Idea summary

When we use the square root property, we always include the \pm symbol to denote the positive and negative root.

We can use the following facts to simplify radical expressions, for a, b \geq 0:

\displaystyle \sqrt{ab}\displaystyle =\displaystyle \sqrt{a} \sqrt{b} Multiplication property of radicals
\displaystyle \sqrt{\dfrac{a}{b}}\displaystyle =\displaystyle \dfrac{\sqrt{a}}{\sqrt{b}}Division property of radicals

Completing the square

Completing the square is a method we use to rewrite a quadratic expression so that it contains a perfect square trinomial which can be factored as A^2+2AB+B^2=\left(A+B\right)^2. We used this method in a previous lesson to convert a quadratic equation from standard form to vertex form. We will now learn to use the completing the square method combined with the square root property to solve quadratic equations.

Exploration

Consider the equation x^2+6x=11.

  1. Try to develop a method for turning the left-hand side of the equation into a perfect square trinomial.
  2. Remember that we need to keep both sides of the equation balanced. After making the perfect square trinomial, check that your equation is still balanced.
  3. How could we solve the equation in this form?
  4. How could you apply your method to x^2-10x-5=0?

For quadratic equations where a=1, we can write them in perfect square form by following these steps:

1\displaystyle x^2+bx+c\displaystyle =\displaystyle 0
2\displaystyle x^2+bx\displaystyle =\displaystyle -cSubtract c from both sides
3\displaystyle x^2+2\left(\frac{b}{2}\right)x\displaystyle =\displaystyle -cRewrite the x term
4\displaystyle x^2+2\left(\frac{b}{2}\right)x+\left(\frac{b}{2}\right)^2\displaystyle =\displaystyle -c+\left(\frac{b}{2}\right)^2Add \left(\dfrac{b}{2}\right)^2 to both sides
5\displaystyle \left(x+\frac{b}{2}\right)^2\displaystyle =\displaystyle -c+\left(\frac{b}{2}\right)^2Factor the perfect square trinomial

If a \neq 1, we can first divide through by a to factor it out.

Note that when we were using completing the square to write an equation in vertex form, we keep the constant term on the same side of the equation as the variable terms. Then, to maintain equivalency and complete square, we add and subtract the same \left({\dfrac{b}{2}}\right)^2 term. This results in all the terms being on the same side of the equation so we can identify the vertex of the parabola.

But, if we want to solve the equation, we keep the x terms together and move the constant term to the other side of the equation. Then the \left({\dfrac{b}{2}}\right)^2 term is added to both sides of the equation to maintain equivalency and create a perfect trinomial. This gives us a squared factor on one side and a constant term on the other side of the equation, allowing us to use the square root property to solve for x. If we can rewrite an equation by completing the square, then we can solve it using square roots.

Examples

Example 4

Solve the following quadratic equations by completing the square.

a

x^{2} + 18 x + 32 = 0

Worked Solution
Create a strategy

To solve an equation by completing the square, start by moving the constant term to the other side of the equation. We will complete the square by finding half the coefficient of the x term, squaring it, and adding it to both sides of the equation. Once we've completed the square, we can solve.

Apply the idea
\displaystyle x^2+18x+32\displaystyle =\displaystyle 0Given equation
\displaystyle x^2+18x\displaystyle =\displaystyle -32Subtract 32 from both sides

Since the coefficient of the x-term is 18, we will need to add \left(\dfrac{18}{2}\right)^2=81 to both sides of our equation.

\displaystyle x^2+18x+81\displaystyle =\displaystyle -32+81Complete the square
\displaystyle (x+9)^2\displaystyle =\displaystyle 49Factor the perfect square trinomial
\displaystyle x+9\displaystyle =\displaystyle \pm 7Take the square root of both sides

This leaves us with two equations x+9=7 and x+9=-7. We can solve both equations by subtracting 9, so we get the solutions x=-2 and x=-16.

b

2x^2 -10x + 7 = 0

Worked Solution
Create a strategy

In this example, the coefficient of x^2 is 2, so we will need to divide this coefficient out before completing the square. We can then perform steps similar to the previous example.

Apply the idea
\displaystyle 2x^2-10x+7\displaystyle =\displaystyle 0Given equation
\displaystyle x^2-5x+\frac{7}{2}\displaystyle =\displaystyle 0Divide both sides by 2
\displaystyle x^2-5x\displaystyle =\displaystyle -\frac{7}{2}Subtract \frac{7}{2} from both sides

The coefficient of x is \dfrac{-10}{2}=-5. Taking half of -5 and squaring it gives us \left(\dfrac{-5}{2}\right)^2=\dfrac{25}{4}, so this is the value that completes the square.

\displaystyle x^2-5x+\dfrac{25}{4}\displaystyle =\displaystyle -\frac{7}{2}+\dfrac{25}{4}Complete the square
\displaystyle \left(x-\dfrac{5}{2}\right)^2\displaystyle =\displaystyle \frac{18}{4}Factor the left side, evaluate the addition on the right side
\displaystyle x-\dfrac{5}{2}\displaystyle =\displaystyle \pm \sqrt{\frac{18}{4}}Square root property
\displaystyle x-\dfrac{5}{2}\displaystyle =\displaystyle \pm \frac{\sqrt{18}}{\sqrt{4}}Division property of radicals
\displaystyle x-\dfrac{5}{2}\displaystyle =\displaystyle \pm \frac{\sqrt{9}\sqrt{2}}{\sqrt{4}}Multiplication property of radicals
\displaystyle x-\dfrac{5}{2}\displaystyle =\displaystyle \pm \frac{3\sqrt{2}}{2}Evaluate the radicals

This leaves us with two equations: x-\dfrac{5}{2}=\dfrac{3\sqrt{2}}{2} and x-\dfrac{5}{2}=-\dfrac{3\sqrt{2}}{2}.

Next, we add \dfrac{5}{2} to solve both equations, and we find that the solutions are {x=\dfrac{5}{2}+\dfrac{3\sqrt{2}}{2}} and {x=\dfrac{5}{2}-\dfrac{3\sqrt{2}}{2}}.

Reflect and check

These can also be combined into one fraction: {x=\dfrac{5\pm3\sqrt{2}}{2}}.

Idea summary

Completing the square can be used to solve any quadratic in the form ax^2+bx+c=0, but it is easiest to use when a=1 and b is even.

Outcomes

A.EI.3

The student will represent, solve, and interpret the solution to a quadratic equation in one variable.

A.EI.3a

Solve a quadratic equation in one variable over the set of real numbers with rational or irrational solutions, including those that can be used to solve contextual problems.

A.EI.3c

Verify possible solution(s) to a quadratic equation in one variable algebraically, graphically, and with technology to justify the reasonableness of answer(s). Explain the solution method and interpret solutions for problems given in context.

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