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8. Solving Quadratic Equations
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United States of AmericaTennessee
Algebra 1

8.03 Solving quadratic equations using square roots

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Lesson

Concept summary

We can solve quadratic equations which are in vertex form by isolating the perfect square and then taking the square root of both sides of the equation.

\displaystyle a\left(x-h\right)^2+k\displaystyle =\displaystyle 0Vertex form
\displaystyle a\left(x-h\right)^2\displaystyle =\displaystyle -kSubtract k
\displaystyle \left(x-h\right)^2\displaystyle =\displaystyle -\frac{k}{a}Divide by -a
\displaystyle x-h\displaystyle =\displaystyle \pm\sqrt{-\frac{k}{a}}Take the square root of both sides
\displaystyle x\displaystyle =\displaystyle h\pm\sqrt{-\frac{k}{a}}Add h

Following these steps, we can see that if -\frac{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 branch. These branches come from the fact that x^2=(-1)^2\cdot x^2=(-x)^2 so \sqrt{x^2}=\pm x.

We can use this solving method for quadratic equations in other forms, by using the completing the square method to first rewrite the equation in vertex form.

Worked examples

Example 1

Solve the following equations by using square roots:

a

x^2=9

Approach

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.

Solution

x=\pm 3

b

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

Approach

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.

Solution

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

c

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

Approach

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

Solution

\displaystyle \left(3x-8\right)^2\displaystyle =\displaystyle 25
\displaystyle 3x-8\displaystyle =\displaystyle \pm 5Square root
\displaystyle 3x\displaystyle =\displaystyle 8\pm 5Add 8
\displaystyle x\displaystyle =\displaystyle \frac{8\pm 5}{3}Divide by 3

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

Example 2

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.

Approach

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.

Solution

\displaystyle (6x)^2\displaystyle =\displaystyle 225
\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.

Reflection

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

Example 3

Consider the following parabola:y = x^{2}-4x+10

a

Determine the minimum or maximum of the parabola.

Approach

The vertex of a parabola will either be the maximum or minimum point, depending on the direction of opening of the parabola.

To find the vertex, we can complete the square to convert the equation from standard form to vertex form.

To determine the direction of opening, we can either graph the parabola, determine another point on the parabola, and/ or look to the scale factor.

Solution

Let's start by coverting our equation to vertex form by completing the square.

Since the coefficient of the x term is -4, we will need to add \left(\dfrac{-4}{2}\right)^2=4 to both sides of our equation. Equivalently, we can add and subtract this amount from one side of the equation.

We'll start with our original equationy = x^{2}-4x+10

\displaystyle y\displaystyle =\displaystyle x^{2}-4x +4 +10 -4Add and subtract 4 on the RHS
\displaystyle y\displaystyle =\displaystyle (x -2 )^{2} +10 -4 Factor the perfect square
\displaystyle y\displaystyle =\displaystyle (x-2)^2 + 6 Simplify

We've completed the square and the equation is now in vertex form. Therefore the vertex is \left(2, 6\right).

Next we'll have to determine whether the vertex is a maximum or minimum.

We can do this by finding another point on the parabola. Let's find the y-intercept by letting x=0.

\displaystyle y\displaystyle =\displaystyle (x-2)^{2} + 6
\displaystyle y\displaystyle =\displaystyle (0-2)^{2} + 6Let x=0
\displaystyle y\displaystyle =\displaystyle 10Simplify

Therefore the y-intercept of the parabola is \left(0, 10\right). Since the point \left(0, 10\right) is above the vertex \left(2, 6\right), it means that our vertex must be the minimum point.

Reflection

We can always determine the direction that a parabola is facing using the scale factor. If the scale factor is positive, then the parabola will be upward facing and the vertex will be a minimum value. If the scale factor is negative, then the parabola will be downward facing and the vertex will be a maximum value.

b

Find the value(s) of x when y=9, if any.

Solution

Since the equation is already in vertex form, we just substitute y=9 into the equation and solve for x by using square roots.

\displaystyle y\displaystyle =\displaystyle (x-2)^2+6Vertex form
\displaystyle 9\displaystyle =\displaystyle (x-2)^2+6Substitute in y=9
\displaystyle 3\displaystyle =\displaystyle (x-2)^2Subtract 6
\displaystyle \pm\sqrt{3}\displaystyle =\displaystyle x-2Take the square root of both sides
\displaystyle 2\pm\sqrt{6}\displaystyle =\displaystyle xAdd 2

When y=9, we have that x=2+\sqrt{3} and x=2-\sqrt{3}.

Outcomes

A1.A.CED.A.1

Create equations and inequalities in one variable and use them to solve problems in a real-world context.*

A1.A.CED.A.2

Create equations in two variables to represent relationships between quantities and use them to solve problems in a real-world context. Graph equations with two variables on coordinate axes with labels and scales, and use the graphs to make predictions.*

A1.A.CED.A.3

Create individual and systems of equations and/or inequalities to represent constraints in a contextual situation, and interpret solutions as viable or non-viable.*

A1.A.CED.A.4

Rearrange formulas to isolate a quantity of interest using algebraic reasoning.

A1.A.REI.A.1

Understand solving equations as a process of reasoning and explain the reasoning. Construct a viable argument to justify a solution method.

A1.A.REI.B.3

Solve quadratic equations and inequalities in one variable.

A1.A.REI.B.3.A

Solve quadratic equations by inspection (e.g., for x^2 = 49), taking square roots, knowing and applying the quadratic formula, and factoring, as appropriate to the initial form of the equation. Recognize when a quadratic equation has solutions that are not real numbers.

A1.MP1

Make sense of problems and persevere in solving them.

A1.MP3

Construct viable arguments and critique the reasoning of others.

A1.MP6

Attend to precision.

A1.MP7

Look for and make use of structure.

A1.MP8

Look for and express regularity in repeated reasoning.

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