Topics
2. Solving Equations
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United States of America
Grades 9-12

2.02 Quadratic equations

Lesson
Practice

Introduction

Solving quadratic equations appeared in Algebra 1, and we can continue to solve quadratic equations using these methods in this lesson. Choosing the most efficient method based on the structure of a quadratic equation is useful when solving.

Solving quadratic equations

Given a quadratic equation, it is helpful to choose the most efficient method for solving the quadratic from our previously learned skills:

Method 1: Using square roots

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 on both sides
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 to both sides

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.

Method 2: Factoring

We can use the zero product property to solve quadratic equations by first writing the equation in factored form: a\left(x-x_1\right)\left(x-x_2\right)=0

If we can write a quadratic equation in the factored form, then we know that either x-x_1=0 or \\ x-x_2=0. This means that the solutions to the quadratic equation are x_1 and x_2. This approach can be useful if the equation has rational solutions.

Method 3: Completing the square

We can use this method to rewrite a quadratic expression so that it contains a perfect square trinomial. A perfect square trinomial takes on the form A^2+2AB+B^2=\left(A+B\right)^2.

If we can rewrite an equation by completing the square, then we can solve it using square roots.

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 0Quadratic equation in standard form
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 coefficient
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.

Method 4: Graphing

If the solutions are integers, drawing the graph of the corresponding quadratic function and finding the x-intercepts is an efficient way to find the solutions.

-4
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-4
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Method 5: Using the quadratic formula

If we are unable to solve the quadratic easily using one of the previusly stated methods, the quadratic formula is often the best approach since it can be used to solve any quadratic equation once it's written in standard form.

Quadratic formula

The formula x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}, used to find solution(s) to a quadratic equation of the form y=ax^2+bx+c

Examples

Example 1

For the following quadratic equations, determine an appropriate strategy for solving, explaining your choice, and then solve for x.

a

-\dfrac{1}{2}x^2+x+12=0

Worked Solution
Create a strategy

If we want to solve this quadratic equation by factoring, we will need the trinomial to have a leading coefficient that is an integer. We can factor out a GCF of -\dfrac{1}{2}, so that the equation becomes -\dfrac{1}{2}(x^2-2x-24)=0.

Since there are no common factors for the remaining three terms, we proceed with finding the value of two integers that multiply to ac = (1)(-24) = -24 and add up to b = -2. After finding these integers, we use them to rewrite the middle term -2x as a sum of two terms.

Lastly, we can factor the trinomial by grouping and solve the equation using the zero product property.

Apply the idea

The factor pair whose sum is -2 is -6 and 4.

We can use this to rewrite the trinomial and factor by grouping as follows:

\displaystyle -\dfrac{1}{2}(x^2-2x-24)\displaystyle =\displaystyle -\dfrac{1}{2}(x^{2} - 6 x +4x - 24)Rewrite polynomial with four terms
\displaystyle =\displaystyle -\dfrac{1}{2}[x(x-6) +4(x-6)]Factor each pair
\displaystyle =\displaystyle -\dfrac{1}{2}(x-6)(x+4)Factor out the GCF of (x-6)

There are no more common factors to be divided out, so the fully factored form of the polynomial is -\dfrac{1}{2}(x-6)(x+4).

This leads to the equation -\dfrac{1}{2}(x-6)(x+4)=0

We can then solve the equation by setting each factor equal to zero, giving us x-6=0 and x+4=0, which gives the solutions x=6 and x=-4.

b

3\left(x-5\right)^2-27 = 0

Worked Solution
Create a strategy

As this is written in vertex form, we can solve it using square roots and inverse operations to solve for x.

Apply the idea
\displaystyle 3\left(x-5\right)^2-27\displaystyle =\displaystyle 0Given equation
\displaystyle 3\left(x-5\right)^2\displaystyle =\displaystyle 27Addition property of equality
\displaystyle \left(x-5\right)^2\displaystyle =\displaystyle 9Division property of equality
\displaystyle x-5\displaystyle =\displaystyle \pm 3Square root property of equality
\displaystyle x\displaystyle =\displaystyle 5\pm 3 Addition property of equality

Giving us two solutions x = 2,\, x = 8

Reflect and check

In general, if we can easily rearrange the equation into the form \left(x-h\right)^2=k for some positive value of k then solving using square roots is a suitable method.

c

3x^2-5x+12=0

Worked Solution
Create a strategy

In general, if the leading coefficient is not 1 then factoring is not likely to be efficient. In this particular case the most efficient method would be to use the quadratic formula. We can also calculate the discriminant b^2-4ac, to identify if there are two, one or no real solutions.

Apply the idea
\displaystyle x\displaystyle =\displaystyle \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}State the quadratic formula
\displaystyle x\displaystyle =\displaystyle \dfrac{-\left(-5\right) \pm \sqrt{\left(-5\right)^2-4\left(3\right)\left(12\right)}}{2\left(3\right)}Substitute values for a,b,c
\displaystyle x\displaystyle =\displaystyle \dfrac{5 \pm \sqrt{-119}}{6}Evaluate the operations

We can see that the discriminant is equal to -119. As it is less than zero, we know that the quadratic equation has no real solutions.

Reflect and check

Using the quadratic formula will always be an appropriate method, and has the advantage of identifying the number and type of solutions, whether they are real or non-real, or rational or non-rational. If you cannot quickly and easily identify a way to solve it using one of the other methods, then using the quadratic formula is always suitable.

d

9x^2-12x-2=0

Worked Solution
Create a strategy

In this example, we would need to factor the trinomial by rewriting it as four terms, where the coefficients of the linear terms have a product of a \cdot c =-18 and a sum of b=-12. Since we do not have factors that add to -12, we can solve the quadratic using the quadratic formula or by completing the square. We can complete the square.

The coefficient of x^2 is 9, so we will need to divide this coefficient out before completing the square.

Once 9 has been divided out, the coefficient of x will be \dfrac{-4}{3}. Taking half of -\dfrac{4}{3} and squaring it gives us \left(\dfrac{-2}{3}\right)^2=\dfrac{4}{9}, so this is the value that completes the square.

Apply the idea
\displaystyle 9x^2-12x-2\displaystyle =\displaystyle 0Given equation
\displaystyle x^2-\dfrac{4}{3}x-\dfrac{2}{9}\displaystyle =\displaystyle 0Divide by 9 on both sides
\displaystyle x^2-\dfrac{4}{3}x\displaystyle =\displaystyle \dfrac{2}{9}Add \dfrac{2}{9} to both sides
\displaystyle x^2-\dfrac{4}{3}x+ \dfrac{4}{9}\displaystyle =\displaystyle \frac{2}{9} + \dfrac{4}{9}Complete the square
\displaystyle \left(x-\dfrac{2}{3}\right)^2\displaystyle =\displaystyle \frac{2}{3}Factor the left side, evaluate the right side
\displaystyle x-\dfrac{2}{3}\displaystyle =\displaystyle \pm \sqrt{\frac{2}{3}}Square root property

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

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

Reflect and check

As stated in part (c), the quadratic formula is always a suitable method. Solving the quadratic equation using the quadratic formula, we have

\displaystyle x\displaystyle =\displaystyle \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}State the quadratic formula
\displaystyle x\displaystyle =\displaystyle \dfrac{-\left(-12\right) \pm \sqrt{\left(-12\right)^2-4\left(9\right)\left(-2\right)}}{2\left(9\right)}Substitute values for a,b,c
\displaystyle x\displaystyle =\displaystyle \dfrac{12 \pm \sqrt{216}}{18}Evaluate the operations

The solutions to the quadratic equation are x= \dfrac{12 + \sqrt{216}}{18} and x= \dfrac{12 - \sqrt{216}}{18}. By simplifying the solutions we found by completing the square and using the quadratic formula, we can confirm that the solutions are equivalent.

Example 2

A sculpture includes a cast iron parabola, coming out of the ground, that reaches a maximum height of 2.25\text{ m}, and has a width of 6\text{ m}.

Let the position of the start of the parabola be \left(0, 0\right). Let x be the horizontal distance and y be the height of the sculpture above the ground.

Determine an appropriate quadratic function that will model the shape of the parabolic sculpture.

Worked Solution
Create a strategy

The vertex lies half way between the two roots and has a height of 2.25 so has coordinates \left(3, 2.25\right). As we know the vertex we will write the function in vertex form f\left(x\right) = a\left(x - h\right)^2 + k. Using another known point we can solve for a. In this case we know that \left(0,0\right) lies on the parabola.

Apply the idea
\displaystyle f\left(x\right)\displaystyle =\displaystyle a\left(x-h\right)^2+kVertex form
\displaystyle f\left(x\right)\displaystyle =\displaystyle a\left(x-3\right)^2+2.25Substitute the vertex \left(3,2.25\right)
\displaystyle 0\displaystyle =\displaystyle a\left(0-3\right)^2+2.25Substitute \left(0,0\right)
\displaystyle 0\displaystyle =\displaystyle 9a+2.25Evaluate the exponent
\displaystyle -2.25\displaystyle =\displaystyle 9aSubtraction property of equality
\displaystyle -\frac{2.25}{9}\displaystyle =\displaystyle aDivision property of equality
\displaystyle a\displaystyle =\displaystyle -\frac{1}{4}Simplify the fraction

The parabola can be modeled by the function f\left(x\right)=-\frac{1}{4}\left(x-3\right)^2+2.25

Reflect and check

As we knew the roots of the parabola, we could have also written the function in factored form, using a similar method. Starting with f\left(x\right)=a\left(x-6\right)x and the substituting in the values of the vertex we would find the equation in factored form is f\left(x\right)=-\dfrac{1}{4}\left(x-6\right)x.

Idea summary

Below is a list of the easiest method to use and the form of the quadratic equation for which we should use it:

Easiest equation form:
Graphing\text{Any form is fine when using technology}
Factoringax^2+bx+c=0\text{ where }a,b,c\text{ are small}
Square root propertyx^2=k\text{ or }a(x-h)^2=k
Completing the squarex^2+bx+c=0\text{ where }b\text{ is even}
Quadratic formulaax^2+bx+c=0\text{ where }a,b,c\text{ are large }

Outcomes

A.SSE.A.2

Use the structure of an expression to identify ways to rewrite it.

A.CED.A.4

Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations.

A.REI.B.4.B

Solve quadratic equations by inspection (e.g. For x^2 = 49), taking square roots, completing the square, the quadratic formula and factoring, as appropriate to the initial form of the equation. Recognize when the quadratic formula gives complex solutions and write them as a ± bi for real numbers a and b.

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