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2. Quadratic Functions & Equations
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United States of AmericaFL
Grade 912

2.02 Solutions to quadratic equations

Lesson
Practice
Lesson

Concept summary

We have several methods we can use to solve quadratic equations. To determine which method is the most suitable we need to look at the form of the quadratic equation. Some are easily solved by factoring for example, or by taking the square root. We may need first need to do a process called completing the square in order to solve by taking square roots.

If we are unable to solve the quadratic easily using one of these 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

If we have access to technology, drawing the graph of the corresponding quadratic function can help us find exact solutions or approximate a solution if it is not an integer value.

As can be seen from different graphs of parabolas, a quadratic equation may have two, one or no real solutions to the equation.

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Worked examples

Example 1

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

a

x^2-2x-24=0

Approach

The leading coefficient of x^2 is 1, so we can check if this equation can be easily factored. The factor pairs of 24 are

  • 1\cdot24
  • 2\cdot12
  • 3\cdot8
  • 4\cdot6

and we want to find two factors that have a product of -24 and sum to -2. As the product is negative we know one factor must be negative and the other positive.

Solution

The two factors that have a product of -24 and a sum of -2 are -6 and 4. We can write the equation in factored form as \left(x-6\right)\left(x+4\right)=0, which gives us two solutions x=6,\, x=-4.

Reflection

In general, if the coefficients are small, and especially if a=1, it is worth checking to see if we can easily factor the equation to solve.

b

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

Approach

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

Solution

\displaystyle 3\left(x-5\right)^2-27\displaystyle =\displaystyle 0
\displaystyle 3\left(x-5\right)^2\displaystyle =\displaystyle 27Add 27
\displaystyle \left(x-5\right)^2\displaystyle =\displaystyle 9Divide by 9
\displaystyle x-5\displaystyle =\displaystyle \pm 3Take the square root
\displaystyle x\displaystyle =\displaystyle 5\pm 3 Add 5

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

Reflection

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

Approach

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.

Solution

\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}Simplify

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.

Reflection

Using the quadratic equation 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.

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 a quadratic function that will model the shape of the parabolic sculpture.

Approach

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.

Solution

\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
\displaystyle 0\displaystyle =\displaystyle a\left(0-3\right)^2+2.25Substitute \left(0,0\right)
\displaystyle 0\displaystyle =\displaystyle 9a+2.25simplify
\displaystyle -2.25\displaystyle =\displaystyle 9aSubtract 2.25
\displaystyle -\frac{2.25}{9}\displaystyle =\displaystyle aDivide by 9
\displaystyle a\displaystyle =\displaystyle -\frac{1}{4}Simplify

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

Reflection

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.

Example 3

A ball is thrown upward and away from the top of a building. The height y (in meters) of the ball at time x (in seconds) is given by: f\left(x\right) = - 3 x^{2} + 12 x + 36

The graph of this relationship is shown.

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a

Identify and interprety the y-intercept of the graph.

Solution

The y-intercept is the height that the ball was thrown from. The ball was thrown from a height of 36\text{ m} . In the context of the question, this is also the height of the building.

Reflection

We can also see from the quadratic function used to model the flight of the ball, which is in standard form, that c=36 which corresponds with the y-intercept.

b

Identify and interpret the x-intercept shown.

Solution

The x-intercept is the time it takes for the ball to reach the ground. The ball reaches the ground after 6 seconds.

Reflection

As the domain is limited to be greater or equal to zero, the negative root can be discounted.

c

Determine the maximum height reached by the ball.

Approach

The maximum height occurs at the vertex of the parabola. As we are given the function in standard form we can find the x-value of the vertex by calculating -\dfrac{b}{2a}. Alternatively, we can read the value from the provided graph.

Solution

The vertex occurs on the line of symmetry which can be found by calculating x=-\dfrac{b}{2a}.

\displaystyle x\displaystyle =\displaystyle -\frac{b}{2a}State the line of symmetry
\displaystyle =\displaystyle -\frac{12}{2\left(-3\right)}Substitute values
\displaystyle =\displaystyle 2Simplify

The line of symmetry is x=2. We can now substitute this value in to y = - 3 x^{2} + 12 x + 36, to find the coordinates of the vertex.

\displaystyle f\left(x\right)\displaystyle =\displaystyle - 3 x^{2} + 12 x + 36State the function
\displaystyle f\left(2\right)\displaystyle =\displaystyle - 3 \left(2\right)^{2} + 12 \left(2\right) + 36Evaluate f\left(2\right)
\displaystyle =\displaystyle 48Simplify

The vertex occurs at \left(2, 48\right), which means the ball reaches a maximum height of 48 meters.

Outcomes

MA.912.AR.3.2

Given a mathematical or real-world context, write and solve one-variable quadratic equations over the real and complex number systems.

MA.912.AR.3.4

Write a quadratic function to represent the relationship between two quantities from a graph, a written description or a table of values within a mathematical or real-world context.

MA.912.AR.3.8

Solve and graph mathematical and real-world problems that are modeled with quadratic functions. Interpret key features and determine constraints in terms of the context.

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