Unit 8 Quadratic Equations — Unit Plan

The height of a softball, in feet, $t$ seconds after someone throws it straight up, can be defined by $f(t) = \text-16t^2+32t+5$. The input of function $f$ is time, and the output is height.

We can find the output of this function at any given input. For instance:

• At the beginning of the softball's journey, when $t = 0$, its height is given by $f(0)$.
• Two seconds later, when $t=2$, its height is given by $f(2)$.

The values of $f(0)$ and $f(2)$ can be found using a graph or by evaluating the expression $\text-16t^2+32t+5$ at those values of $t$. What if we know the output of the function and want to find the inputs? For example:

• When does the softball hit the ground?

  Answering this question means finding the values of $t$ that make $f(t)=0$, or solving $\text -16t^2+32t+5=0$. 

• How long will it take the ball to reach 8 feet?

  This means finding one or more values of $t$ that make $f(t) = 8$, or solving the equation $\text -16t^2+32t+5=8$.

The equations $\text -16t^2+32t+5=0$ and $\text -16t^2+32t+5=8$ are quadratic equations. One way to solve these equations is by graphing $y = f(t)$.

• To answer the first question, we can look for the horizontal intercepts of the graph, where the vertical coordinate is 0.
• To answer the second question, we can look for the horizontal coordinates that correspond to a vertical coordinate of 8.

We can see that there are two solutions to the equation $\text -16t^2+32t+5=8$ and one solution to the equation $\text{-}16t^2+32t+5=0$.

The softball has a height of 8 feet twice, when going up and when coming down, and these instances occur when $t$ is about 0.1 and 1.9. It has a height of 0 once, when $t$ is about 2.15.

Often, when we are modeling a situation mathematically, an approximate solution is good enough. Sometimes, however, we would like to know exact solutions, and it may not be possible to find them using a graph. In this unit, we will learn more about quadratic equations and how to solve for exact answers using algebraic techniques.

The height of a potato that is launched from a mechanical device can be modeled by a function, $g$, with $x$ representing time in seconds. Here are two expressions that are equivalent and both define function $g$.

Notice that one expression is in standard form and the other is in factored form.

Suppose we wish to know, without graphing the function, the time when the potato will hit the ground. We know that the value of the function at that time is 0, so we can write:

Let's try solving $\text-16x^2+80x+96 = 0$, using some familiar moves. For example:

These steps don’t seem to get us any closer to a solution. We need some new moves!

What if we use the other equation? Can we find the solutions to $\text-16(x-6)(x+1)=0$?

Earlier, we learned that the zeros of a quadratic function can be identified when the expression defining the function is in factored form. The solutions to $\text-16(x-6)(x+1)=0$ are the zeros to function $g$, so this form may be more helpful! We can reason that:

• If $x$ is 6, then the value of $x-6$ is 0, so the entire expression has a value of 0.
• If $x$ is -1, then the value of $x+1$ is 0, so the entire expression also has a value of 0.

This tells us that 6 and -1 are solutions to the equation, and that the potato hits the ground after 6 seconds. (A negative value of time is not meaningful, so we can disregard the -1.)

Both equations we see here are quadratic equations. In general, a quadratic equation is an equation that can be expressed as $ax^2+bx+c=0$, where $a$, $b$, and $c$ are constants and $a \neq 0$.

In upcoming lessons, we will learn how to rewrite quadratic equations into forms that make the solutions easy to see.

Some quadratic equations can be solved by performing the same operation on each side of the equal sign and reasoning about which values for the variable would make the equation true.

Suppose we wanted to solve $3(x+1)^2-75=0$. We can proceed like this:

• If $x+1 = 5$, then $x=4$.
• If $x+1 = \text-5$, then $x=\text-6$.

This means that both $x=4$ and $x=\text-6$ make the equation true and are solutions to the equation.

Many quadratic equations have 2 solutions, but some have only 1 or no solution.

The zero product property says that if the product of two numbers is 0, then one of the numbers must be 0. In other words, if $a\boldcdot b=0,$ then either $a=0$ or $b=0$. This property is handy when an equation we want to solve states that the product of two factors is 0.

Suppose we want to solve $m(m+9)=0$. This equation says that the product of $m$ and $(m+9)$ is 0. For this to be true, either $m=0$ or $m+9=0$, so both 0 and -9 are solutions.

Here is another equation: $(u-2.345)(14u+2)=0$. The equation says the product of $(u-2.345)$ and $(14u+2)$ is 0, so we can use the zero product property to help us find the values of $u$. For the equation to be true, one of the factors must be 0.

• For $u-2.345=0$ to be true, $u$ would have to be 2.345.
• For $14u+2=0$ or ($14u = \text-2$) to be true, $u$ would have to be $\text-\frac{2}{14}$, or $\text-\frac17$.

The solutions are 2.345 and $\text-\frac17$.

In general, when a quadratic expression in factored form is on one side of an equation and 0 is on the other side, we can use the zero product property to find its solutions.

This property is unique to 0. Given an equation like $a \boldcdot b = 6$, the factors could be 2 and 3, 1 and 6, -12 and $\text{-}\frac{1}{2}$, $\pi$ and $\frac{6}{\pi}$, or any other of the infinite number of combinations. This type of equation does not give insight into the value of $a$ or $b$.

Previously, you learned how to expand a quadratic expression in factored form and write it in standard form by applying the distributive property.

For example, to expand $(x+4)(x+5)$, we apply the distributive property to multiply $x$ by $(x+5)$ and 4 by $(x+5)$. Then, we apply the property again to multiply $x$ by $x$, $x$ by 5, 4 by $x$, and 4 by 5. 

The diagram helps us see that $(x+4)(x+5)$ is equivalent to $x^2 +5x +4x + 4 \boldcdot 5$, or in standard form, $x^2 +9x + 20$.

• The linear term, or the term with a single factor of $x$ in the standard form of a quadratic expression, is $9x$ and has a coefficient of 9, which is the sum of 5 and 4.
• The constant term, 20, is the product of 5 and 4.

We can use these observations to reason in the other direction: starting with an expression in standard form and writing it in factored form.

For example, suppose we wish to write $x^2 - 11x + 24$ in factored form.

When we rewrite expressions in factored form, it is helpful to remember that:

• Multiplying two positive numbers or two negative numbers results in a positive product.
• Multiplying a positive number and a negative number results in a negative product.

This means that if we want to find two factors whose product is 10, the factors must both be positive or both be negative. If we want to find two factors whose product is -10, one of the factors must be positive and the other negative.

Suppose we wanted to rewrite $x^2 -8x + 7$ in factored form. Recall that subtracting a number can be thought of as adding the opposite of that number, so that expression can also be written as $x^2 + \text-8x + 7$. We are looking for two numbers that:

• Have a product of 7. The candidates are 7 and 1, and -7 and -1.
• Have a sum of -8. Only -7 and -1 from the list of candidates meet this condition.

The factored form of $x^2 -8x + 7$ is therefore $(x + \text-7)(x + \text-1)$ or, written another way, $(x-7)(x-1)$.

To write $x^2 + 6x - 7$ in factored form, we would need two numbers that:

• Multiply to make -7. The candidates are 7 and -1, and -7 and 1.
• Add up to 6. Only 7 and -1 from the list of candidates add up to 6.

The factored form of $x^2 + 6x - 7$ is $(x+7)(x-1)$.

Recently, we learned strategies for transforming expressions from standard form to factored form. In earlier lessons, we have also seen that when a quadratic expression is in factored form, we can find values of the variable that make the expression equal zero. Suppose we are solving the equation $x(x+4)=0$, which says that the product of $x$ and $x+4$ is 0. By the zero product property, we know this means that either $x=0$ or $x+4=0$, which then tells us that 0 and -4 are solutions.

Together, these two skills—writing quadratic expressions in factored form and using the zero product property when a factored expression equals 0—allow us to solve quadratic equations given in other forms. Here is an example:

$\displaystyle \begin{aligned} n^2-4n &= 140 &\qquad& \text{Original equation}\\n^2-4n-140 & =0 &\qquad& \text{Subtract 140 from each side so the right side is 0.}\\ (n-14)(n+10) &= 0 &\qquad& \text{Rewrite in factored form.} \\ \\n-14=0 \quad \text{or} \quad &n+10=0 &\qquad& \text {Apply the zero product property.} \\ n=14 \quad \text{or} \quad &n=\text-10 &\qquad& \text{Solve each equation.}\end{aligned}$

When a quadratic equation is written as as an expression in factored form equal to 0, we can also see the number of solutions the equation has.

In the previous example, it is not obvious how many solutions there are when the equation is in the form $n^2-4n-140=0$. When the equation is rewritten as $(n-14)(n+10) =0$, we can see that there are two numbers that could make the expression equal 0: 14 and -10.

How many solutions does the equation $x^2-20x+100=0$ have?

Let’s rewrite it in factored form: $(x-10)(x-10)=0$. The two factors are identical, which means that there is only one value of $x$ that makes the expression $(x-10)(x-10)$ equal 0. The equation has only one solution: 10.

Turning an expression into a perfect square can be a good way to solve a quadratic equation. Suppose we wanted to solve $x^2 - 14x +10 = \text-30$.

The expression on the left, $x^2 - 14x +10$, is not a perfect square, but $x^2 - 14x + 49$ is a perfect square. Let’s transform that side of the equation into a perfect square (while keeping the equality of the two sides).

This method of solving quadratic equations is called completing the square. In general, perfect squares in standard form look like $x^2 + bx + \left(\frac{b}{2} \right)^2$, so to complete the square, take half of the coefficient of the linear term and square it.

In the example, half of -14 is -7, and $(\text-7)^2$ is 49. We wanted to make the left side $x^2 - 14x + 49.$ To keep the equation true and maintain equality of the two sides of the equation, we added 49 to each side.

It takes some practice to become proficient at completing the square, but it makes it possible to solve many more equations than we could by methods we learned previously.

We have learned a couple of methods for solving quadratic equations algebraically:

• By rewriting the equation as so that one side is 0 and the other side is in factored form, then using the zero product property
• By completing the square

Some equations can be solved quickly with one of these methods, but many cannot. Here is an example: $5x^2-3x-1=0$. The expression on the left cannot be rewritten in factored form with rational coefficients. Because the coefficient of the squared term is not a perfect square, and the coefficient of the linear term is an odd number, completing the square would be inconvenient and would result in a perfect square with fractions.

The quadratic formula can be used to find the solutions to any quadratic equation, including those that are tricky to solve with other methods.

For the equation $5x^2-3x-1=0$, we see that $a=5$, $b=\text-3$, and $c=\text-1$. Let’s solve it!

$\displaystyle \begin{aligned} x &=\dfrac{\text- b \pm \sqrt{b^2-4ac}}{2a} &\qquad &\text{the quadratic formula}\\ x &=\dfrac{\text-(\text-3) \pm \sqrt{(\text-3)^2-4(5)(\text-1)}}{2(5)} &\qquad &\text{Substitute the values of }a, b, \text{and }c. &\\ x &=\dfrac{3 \pm \sqrt{9+20}}{10} &\qquad &\text {Evaluate each part of the expression.} \\ x &=\dfrac{3 \pm \sqrt{29}}{10}\end{aligned}$

A calculator gives approximate solutions of 0.84 and -0.24 for $\frac{3 + \sqrt{29}}{10}$ and $\frac{3 - \sqrt{29}}{10}$.

We can also use the formula for simpler equations like $x^2-9x+8=0$, but it may not be the most efficient way. If the quadratic expression can be easily rewritten in factored form or made into a perfect square, those methods may be preferable. For example, rewriting $x^2-9x+8=0$ as $(x-1)(x-8)=0$ immediately tells us that the solutions are 1 and 8.

Quadratic equations that represent situations cannot always be neatly put into factored form or easily solved by finding square roots. Completing the square is a workable strategy, but for some equations, it may involve many cumbersome steps. Graphing is also a handy way to solve the equations, but it doesn’t always give us precise solutions.

With the quadratic formula, we can solve these equations more readily and precisely.

Here’s an example: Function $h$ models the height of an object, in meters, $t$ seconds after it is launched into the air. It is is defined by $h(t)=\text-5t^2+25t$.

To know how much time it would take the object to reach 15 meters, we could solve the equation $15=\text-5t^2+25t$. How should we do it?

• Rewriting it in standard form gives $\text-5t^2+25t-15=0$. The expression on the left side of the equation cannot be written in factored form, however.
• Completing the square isn't convenient because the coefficient of the squared term is not a perfect square and the coefficient the linear term is an odd number.
• Let’s use the quadratic formula, using $a=\text-5,b=25,\text{and }c=\text-15$!

The solutions tell us that there are two times after the launch when the object is at a height of 15 meters: at about 0.7 second (as the object is going up) and 4.3 seconds (as it comes back down).

Remember that a quadratic function can be defined by equivalent expressions in different forms, which enable us to see different features of its graph. For example, these expressions define the same function:

$\begin{aligned} (x-3)(x-7) \qquad &\text {factored form}\\ x^2-10x+21 \qquad &\text {standard form}\\ (x-5)^2-4 \qquad &\text {vertex form}\end{aligned}$

• From factored form, we can tell that the $x$-intercepts are $(3,0)$ and $(7,0)$.
• From standard form, we can tell that the $y$-intercept is $(0,21)$.
• From vertex form, we can tell that the vertex is $(5,\text-4)$.

Graph of a quadratic function on a grid. X axis from 0 to 9, by 1’s. Y axis from negative 8 to 20, by 4’s. Origin O. Graph opens upward with points 0 comma 21, 3 comma 0, and 7 comma 0. Vertex at 5 comma negative 4.

Recall that a function expressed in vertex form is written as $a(x-h)^2 + k$. The values of $h$ and $k$ reveal the vertex of the graph: $(h, k)$ are the coordinates of the vertex. In this example, $a$ is 1, $h$ is 5, and $k$ is -4.

• If we have an expression in vertex form, we can rewrite it in standard form by using the distributive property and combining like terms.

  Let’s say we want to rewrite $(x-1)^2-4$ in standard form.

  $\displaystyle \begin{aligned} &(x-1)^2-4 \\ &(x-1)(x-1)-4 \\& x^2-2x+1-4 \\&x^2-2x-3 \end{aligned}$

• If we have an expression in standard form, we can rewrite it in vertex form by completing the square.

  Let’s rewrite $x^2 + 10x +24$ in vertex form.

  A perfect square would be $x^2 + 10x +25$, so we need to add 1. Adding 1, however, would change the expression. To keep the new expression equivalent to the original one, we will need to both add 1 and subtract 1.

  $\displaystyle \begin{aligned} &x^2 + 10x +24 \\& x^2 + 10x +24 + 1 - 1 \\ &x^2 + 10x + 25 - 1 \\ &(x+5)^2-1 \end{aligned}$

• Let’s rewrite another expression in vertex form: $\text-2x^2 + 12x - 30$.

  To make it easier to complete the square, we can use the distributive property to rewrite the expression with -2 as a factor, which gives $\text-2 (x^2 -6x + 15)$.

  For the expression in the parentheses to be a perfect square, we need $x^2 -6x + 9$. We have 15 in the expression, so we can subtract 6 from it to get 9, and then add 6 again to keep the value of the expression unchanged. Then, we can rewrite $x^2-6x+9$ in factored form.

  $\displaystyle \begin{aligned} &\text-2x^2 + 12x -30 \\&\text-2(x^2 -6x +15) \\& \text-2(x^2 -6x +15 -6 +6) \\ &\text-2(x^2 -6x +9 +6) \\ &\text-2\left((x-3)^2 + 6\right) \end{aligned}$

  This expression is not yet in vertex form, however. To finish up, we need to apply the distributive property again so that the expression is of the form $a(x-h)^2 + k$:

  $\displaystyle \begin{aligned} &\text-2\left((x-3)^2 + 6\right) \\&\text-2(x-3)^2 -12 \end{aligned}$

  When written in this form, we can see that the vertex of the graph representing $\text-2(x-3)^2 -12$ is $(3, \text-12)$.

Any quadratic function has either a maximum or a minimum value. We can tell whether a quadratic function has a maximum or a minimum by observing the vertex of its graph.

Here are graphs representing functions $f$ and $g$, defined by $f(x)=\text-(x+5)^2+4$ and $g(x)=x^2+6x-1$.

• The vertex of the graph of $f$ is $(\text-5,4)$, and the graph is a parabola that opens downward.
• No other points on the graph of $f$ (no matter how much we zoom out) are higher than $(\text-5,4)$, so we can say that $f$ has a maximum of 4, and that this occurs when $x=\text-5$.

• The vertex of the graph of $g$ is at $(\text-3, \text-10)$, and the graph is a parabola that opens upward.
• No other points on the graph (no matter how much we zoom out) are lower than $(\text-3,\text-10)$, so we can say that $g$ has a minimum of -10, and that this occurs when $x = \text-3$.

We know that a quadratic expression in vertex form can reveal the vertex of the graph, so we don’t actually have to graph the expression. But how do we know, without graphing, if the vertex corresponds to a maximum or a minimum value of a function?

The vertex form can give us that information as well!

To see if $(\text-3, \text-10)$ is a minimum or maximum of $g$, we can rewrite $x^2+6x-1$ in vertex form, which is $(x+3)^2-10$. Let’s look at the squared term in $(x+3)^2-10$.

• When $x=\text-3$, $(x+3)$ is 0, so $(x+3)^2$ is also 0.
• When $x$ is not -3, the expression $(x+3)$ is a nonzero number, and $(x+3)^2$ is positive.
• Because a squared number cannot have a value less than 0, $(x+3)^2$ has the least value when $x=\text{-}3$.

To see if $(\text-5,4)$ is a minimum or maximum of $f$, let’s look at the squared term in $\text-(x+5)^2+4$.

• When $x=\text-5$, $(x+5)$ is 0, so $(x+5)^2$ is also 0.
• When $x$ is not -5, the expression $(x+5)$ is nonzero, so $(x+5)^2$ is positive. The expression $\text-(x+5)^2$ has a coefficient of -1, however. Multiplying $(x+5)^2$ (which is positive when $x \neq \text-5$) by a negative number results in a negative number.
• Because a negative number is always less than 0, the value of $\text-(x+5)^2+4$ will always be less when $x \neq \text-5$ than when $x=\text-5$. This means $x =\text-5$ gives the greatest value of $f$.