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Study Guides > Intermediate Algebra

Quadratic Equations

Learning Outcomes

  • Quadratic Equations
    • Recognize a quadratic equation
    • Use the zero product principle to solve a quadratic equation that can be factored
    • Determine when solutions to quadratic equations can be discarded
  • Pythagorean Theorem
    • Recognize a right triangle from other types of triangles
    • Use the Pythagorean theorem to find the lengths of a right triangle
  • Projectiles
    • Define projectile motion
    • Solve a quadratic equation that represents projectile motion
    • Interpret the solution to a quadratic equation that represents projectile motion
When a polynomial is set equal to a value (whether an integer or another polynomial), the result is an equation. An equation that can be written in the form ax2+bx+c=0ax^{2}+bx+c=0 is called a quadratic equation. You can solve a quadratic equation using the rules of algebra, applying factoring techniques where necessary, and by using the Principle of Zero Products. There are many applications for quadratic equations. When you use the Principle of Zero Products to solve a quadratic equation, you need to make sure that the equation is equal to zero. For example, 122+11x+2=712^{2}+11x+2=7 must first be changed to 12x2+11x+−5=012x^{2}+11x+-5=0 by subtracting 7 from both sides.

Example

The area of a rectangular garden is 30 square feet. If the length is 7 feet longer than the width, find the dimensions.

Answer: The formula for the area of a rectangle is Area=length⋅width\text{Area}=\text{length}\cdot\text{width}, or A=l⋅wA=l\cdot{w}.

            A=l⋅w width=wlength=w+7     area=3030=(w+7)(w)\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,A=l\cdot{w}\\\,\text{width}=w\\\text{length}=w+7\\\,\,\,\,\,\text{area}=30\\\\30=\left(w+7\right)\left(w\right)\end{array}

Multiply.

30=w2+7w30=w^{2}+7w

Subtract 30 from both sides to set the equation equal to 0.

w2+7w–30=0w^{2}+7w–30=0

Find two numbers whose product is −30−30 and whose sum is 7, and write the middle term as 10w–3w10w–3w.

w2+10w–3w–30=0w^{2}+10w–3w–30=0

Factor w out of the first pair and −3−3 out of the second pair.

w(w+10)−3(w+10)=0w\left(w+10\right)-3\left(w+10\right)=0

Factor out w+10w+10.

(w–3)(w+10)=0\left(w–3\right)\left(w+10\right)=0

Use the Zero Product Property to solve for w.

w−3=0      or      w+10=0       w=3                        w=−10\begin{array}{l}w-3=0\,\,\,\,\,\,\text{or}\,\,\,\,\,\,w+10=0\\\,\,\,\,\,\,\,w=3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,w=-10\end{array}

The solution w=−10w=−10 does not work for this application, as the width cannot be a negative number, we discard the −10−10. So, the width is 3 feet.

The width = 3 feet

Substitute w=3w=3 into the expression w+7w+7 to find the length: 3+7=103+7=10.

The length is 3+7=103+7=10 feet

Answer

The width of the garden is 3 feet, and the length is 10 feet.

In the example in the following video, we present another area application of factoring trinomials. https://youtu.be/PvXsWZp588o The example below shows another quadratic equation where neither side is originally equal to zero. (Note that the factoring sequence has been shortened.)

Example

Solve 5b2+4=−12b5b^{2}+4=−12b for b.

Answer: The original equation has −12b−12b on the right. To make this side equal to 0, add 12b12b to both sides.

5b2+4+12b=−12b+12b5b^{2}+4+12b=−12b+12b

Combine like terms.

5b2+12b+4=05b^{2}+12b+4=0

Rewrite 12b12b as 10b+2b10b+2b.

5b2+10b+2b+4=05b^{2}+10b+2b+4=0

Factor out 5b5b from the first pair and 2 from the second pair.

5b(b+2)+2(b+2)=05b\left(b+2\right)+2\left(b+2\right)=0

Factor out b+2b+2.

(5b+2)(b+2)=0\left(5b+2\right)\left(b+2\right)=0

Apply the Zero Product Property.

5b+2=0   or   b+2=05b+2=0\,\,\,\text{or}\,\,\,b+2=0

Solve each equation.

b=−25   OR   b=−2b=-\frac{2}{5}\,\,\,\text{OR}\,\,\,b=−2

Answer

b=−25   or   b=−2b=-\frac{2}{5}\,\,\,\text{or}\,\,\,b=−2

The following video contains another example of solving a quadratic equation using factoring with grouping. https://youtu.be/04zEXaOiO4U   If you factor out a constant, the constant will never equal 0. So it can essentially be ignored when solving. See the following example.

Example

Solve for k: −2k2+90=−8k-2k^2+90=-8k

Answer: We need to move all the terms to one side so we can use the zero product principle.

−2k2+90=−8k+8k‾                +8k‾−2k2+8k+90=0\begin{array}{l}-2k^2+90=-8k\\\underline{+8k}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{+8k}\\-2k^2+8k+90=0\end{array}

You will either need to try to factor out a -2, or use the method where we multiply −2⋅90-2\cdot{90} and find factors that sum to 8. Each term is divisible by 2, so we can factor out -2.

−2(k2−4k−45)=0-2\left(k^2-4k-45\right)=0

Note how we changed the signs when we factored out a negative number. If we can factor the polynomial, we will be able to solve.   Using the shortcut for factoring we will start with the variable and place a plus and a minus sign in the binomials. We do this because 45 is negative and the only way to get a product that is negative is if one of the factors is negative.

−2(k−   )(k+   )=0-2\left(k-\,\,\,\right)\left(k+\,\,\,\right)=0

We want our factors to have a product of -45 and a sum of -4:

Factors whose product is -45 Sum of the factors
1⋅−45=−451\cdot-45=-45 1−45=−441-45=-44
3⋅−15=−453\cdot-15=-45 3−15=−123-15=-12
5⋅−9=−455\cdot-9=-45 5−9=−45-9=-4
There are more factors that will give -45, but we have found the ones that sum to -4, so we will stop. Fill in the rest of the binomials with the factors we found.

−2(k−9)(k+5)=0-2\left(k-9\right)\left(k+5\right)=0

Now we can set each factor equal to zero using the zero product rule.

−2=0-2=0 This solution is nonsense so we discard it.

k−9=0,k=9k-9=0, k=9

k+5=0,k=−5k+5=0, k=-5

Answers

k=9 and k=−5k=9\text{ and }k=-5

In this last video example, we solve a quadratic equation with a leading coefficient of -1 using the shortcut method of factoring and the zero product principle. https://youtu.be/nZYfgHygXis

Pythagorean Theorem

Four types of triangles, scalene, right, equaliateral, and isosceles. Triangles
The Pythagorean theorem or Pythagoras's theorem is a statement about the sides of a right triangle. One of the angles of a right triangle is always equal to 90 degrees. This angle is the right angle. The two sides next to the right angle are called the legs and the other side is called the hypotenuse. The hypotenuse is the side opposite to the right angle, and it is always the longest side. The image above shows four common kinds of triangle, including a right triangle.
right triangle labeled with teh longest length = a, and the other two b and c. Right Triangle Labeled

The Pythagorean theorem is often used to find unknown lengths of the sides of right triangles. If the longest leg of a right triangle is labeled c, and the other two a, and b as in the image on teh left,  The Pythagorean Theorem states that

a2+b2=c2a^2+b^2=c^2

Given enough information, we can solve for an unknown length.  This relationship has been used for many, many years for things such as celestial navigation and early civil engineering projects. We now have digital GPS and survey equipment that have been programmed to do the calculations for us.

In the next example we will combine the power of the Pythagorean theorem and what we know about solving quadratic equations to find unknown lengths of right triangles.

Example

A right triangle has one leg with length x, another whose length is greater by two,  and the length of the hypotenuse is greater by four.  Find the lengths of the sides of the triangle. Use the image below. Right triangle with one leg having length = x, one with length= x+2 and the hypotenuse = x+4

Answer: Read and understand: We know the lengths of all the sides of a triangle in terms of one side. We also know that the Pythagorean theorem will give us a relationship between the side lengths of a right triangle. Translate: 

a2+b2=c2x2+(x+2)2=(x+4)2\begin{array}{l}a^2+b^2=c^2\\x^2+\left(x+2\right)^2=\left(x+4\right)^2\end{array}

Solve:  If we can move all the terms to one side and factor, we can use the zero product principle to solve.  Since this is the only method we know - let's hope it works! First, multiply the binomials and simplify so we can see what we are working with.

x2+(x+2)2=(x+4)2x2+x2+4x+4=x2+8x+162x2+4x+4=x2+8x+16\begin{array}{l}x^2+\left(x+2\right)^2=\left(x+4\right)^2\\x^2+x^2+4x+4=x^2+8x+16\\2x^2+4x+4=x^2+8x+16\end{array}

Now move all the terms to one side and see if we can factor.

2x2+4x+4=x2+8x+16−x2‾   −8x‾   −16‾     −x2‾   −8x‾   −16‾x2−4x−12=0\begin{array}{l}2x^2+4x+4=x^2+8x+16\\\underline{-x^2}\,\,\,\underline{-8x}\,\,\,\underline{-16}\,\,\,\,\,\underline{-x^2}\,\,\,\underline{-8x}\,\,\,\underline{-16}\\x^2-4x-12=0\end{array}

This went from a messy looking problem to something promising. We can factor using the shortcut:

−6⋅2=−12, and −6+2=−4-6\cdot{2}=-12,\text{ and }-6+2=-4

So we can build our binomial factors with -6 and 2:

(x−6)(x+2)=0\left(x-6\right)\left(x+2\right)=0

Set each factor equal to zero:

x−6=0,x=6x-6=0, x=6

x+2=0,x=−2x+2=0, x=-2

Interpret: Ok, it doesn't make sense to have a length equal to -2, so we can safely throw that solution out.  The lengths of the sides are as follows:

x=6x=6

x+2=6+2=8x+2=6+2=8

x+4=6+4=10x+4=6+4=10

Check: Since we know the relationship between the sides of a right triangle we can check that we are correct. Sometimes it helps to draw a picture

.Screen Shot 2016-06-14 at 9.20.06 PM

We know that a2+b2=c2a^2+b^2=c^2, so we can substitute the values we found:

62+82=10236+64=100100=100\begin{array}{l}6^2+8^2=10^2\\36+64=100\\100=100\end{array}

Our solution checks out.

Answer

The lengths of the sides of the right triangle are 6, 8, and 10

This video example shows another way a quadratic equation can be used to find and unknown length of a right triangle. https://youtu.be/xeP5pRBqsNs If you are interested in celestial navigation and the mathematics behind it, watch this video for fun. https://www.youtube.com/watch?v=XWLZKmPU17M

Projectile Motion

Projectile motion happens when you throw a ball into the air and it comes back down because of gravity.  A projectile will follow a curved path that behaves in a predictable way.  This predictable motion has been studied for centuries, and in simple cases it's height from the ground at a given time, t, can be modeled with a quadratic polynomial of the form height=at2+bt+c\text{height}=at^2+bt+c such as we have been studying in this module. Projectile motion is also called a parabolic trajectory because of the shape of the path of a projectile's motion, as in the image of water in the fountain below.
Water from a fountain shoing classic parabolic motion. Parabolic WaterTrajectory
Parabolic motion and it's related equations allow us to launch satellites for telecommunications, and rockets for space exploration. Recently, police departments have even begun using projectiles with GPS to track fleeing suspects in vehicles, rather than pursuing them by high-speed chase [footnote]"Cops' Latest Tool in High-speed Chases: GPS Projectiles." CBSNews. CBS Interactive, n.d. Web. 14 June 2016.[/footnote]. In this section we will solve simple quadratic polynomials that represent the parabolic motion of a projectile. The real mathematical model for the path of a rocket or a police GPS projectile may have different coefficients or more variables, but the concept remains the same. We will also learn to interpret the meaning of the variables in a polynomial that models projectile motion.

Example

A small toy rocket is launched from a 4-foot pedestal. The height (h, in feet) of the rocket t seconds after taking off is given by the formula h=−2t2+7t+4h=−2t^{2}+7t+4. How long will it take the rocket to hit the ground?

Answer: Read and understand: The rocket will be on the ground when the height is 0. We want to know how long, t,  the rocket is in the air. Translate: So, we will substitute 0 for h in the formula and solve for t.

h=−2t2+7t+40=−2t2+7t+4\begin{array}{l}h=−2t^{2}+7t+4\\0=−2t^{2}+7t+4\end{array}

Write and Solve: Rewrite the middle term using the a⋅ca\cdot{c} method.

0=−2t2+8t−t+40=-2t^{2}+8t-t+4

Factor the trinomial by grouping.

0=−2t(t−4)−(t−4)0=(−2t−1)(t−4)0=−1(2t+1)(t−4)\begin{array}{l}0=-2t\left(t-4\right)-\left(t-4\right)\\0=\left(-2t-1\right)\left(t-4\right)\\0=-1\left(2t+1\right)\left(t-4\right)\end{array}

Use the Zero Product Property. There is no need to set the constant factor −1-1 to zero, because −1-1 will never equal zero.

2t+1=0      or      t−4=02t+1=0\,\,\,\,\,\,\text{or}\,\,\,\,\,\,t-4=0

Solve each equation.

t=−12      or      t=4t=-\frac{1}{2}\,\,\,\,\,\,\text{or}\,\,\,\,\,\,t=4

Interpret the answer. Since t represents time, it cannot be a negative number; only t=4t=4 makes sense in this context. t=4t=4

Answer

The rocket will hit the ground 4 seconds after being launched.

In the next example we will solve for the time that the rocket is at a given height other than zero.

Example

Use the formula for the height of the rocket in the previous example to find the time when the rocket is 4 feet from hitting the ground on it's way back down.  Refer to the image. h=−2t2+7t+4h=−2t^{2}+7t+4 Parabolic motion of rocket which starts four feet up from the ground. t=0 is labeled at the starti of hte parabolic motion adn t=? is labeled at four feet from the ground on the other side of the parabola.

Answer: Read and understand: We are given that the height of the rocket is 4 feet from the ground on it's way back down. We want to know how long it has taken the rocket to get to that point in it's path, we are going to solve for t. Translate: Substitute h = 4 into the formula for height, and try to get zero on one side since we know we can use the zero product principle to solve polynomials. Write and Solve:

h=−2t2+7t+44=−2t2+7t+4−4‾                             −4‾0=−2t2+7t\begin{array}{l}h=−2t^{2}+7t+4\\4=-2t^2+7t+4\\\underline{-4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-4}\\0=-2t^2+7t\end{array}

Now we can factor out a t from each term:

0=t(−2t+7)0=t\left(-2t+7\right)

Solve each equation for t using the zero product principle:

t=0 OR −2t+7=0                                      −7‾       −7‾                                      −2t−2=−7−2                                      t=72=3.5\begin{array}{l}t=0\text{ OR }-2t+7=0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-7}\,\,\,\,\,\,\,\underline{-7}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{-2t}{-2}=\frac{-7}{-2}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t=\frac{7}{2}=3.5\end{array}

Interpret: It doesn't make sense for us to choose t=0 because we are interested in the amount of time that has passed when the projectile is 4 feet from hitting the ground on it's way back down. We will choose t=3.5

Answer

t=3.5 seconds t=3.5\text{ seconds }

The video that follows presents another example of solving a quadratic equation that represents parabolic motion. https://youtu.be/hsWSzu3KcPU In this section we introduced the concept of projectile motion, and showed that it can be modeled with a quadratic polynomial.  While the models used in these examples are simple, the concepts and interpretations are the same.  The methods used to solve quadratic polynomials that don't factor easily are many and well known, it is likely you will come across more in your studies.

Summary

You can find the solutions, or roots, of quadratic equations by setting one side equal to zero, factoring the polynomial, and then applying the Zero Product Property. The Principle of Zero Products states that if ab=0ab=0, then either a=0a=0 or b=0b=0, or both a and b are 0. Once the polynomial is factored, set each factor equal to zero and solve them separately. The answers will be the set of solutions for the original equation. Not all solutions are appropriate for some applications. In many real-world situations, negative solutions are not appropriate and must be discarded.

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