8.6: Polynomial Equations
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Learning Objectives
By the end of this section, you will be able to:
- Use the Zero Product Property
- Solve quadratic equations by factoring
- Solve equations with polynomial functions
- Solve applications modeled by polynomial equations
Before you get started, take this readiness quiz.
We have spent considerable time learning how to factor polynomials. We will now look at polynomial equations and solve them using factoring, if possible.
A polynomial equation is an equation that contains a polynomial expression. The degree of the polynomial equation is the degree of the polynomial.
POLYNOMIAL EQUATION
A polynomial equation is an equation that contains a polynomial expression.
The degree of the polynomial equation is the degree of the polynomial.
We have already solved polynomial equations of degree one . Polynomial equations of degree one are linear equations are of the form \(ax+b=c\).
We are now going to solve polynomial equations of degree two . A polynomial equation of degree two is called a quadratic equation . Listed below are some examples of quadratic equations:
\[x^2+5x+6=0 \qquad 3y^2+4y=10 \qquad 64u^2−81=0 \qquad n(n+1)=42 \nonumber\]
The last equation doesn’t appear to have the variable squared, but when we simplify the expression on the left we will get \(n^2+n\).
The general form of a quadratic equation is \(ax^2+bx+c=0\), with \(a\neq 0\). (If \(a=0\), then \(0·x^2=0\) and we are left with no quadratic term.)
QUADRATIC EQUATION
An equation of the form \(ax^2+bx+c=0\) is called a quadratic equation.
\[a,b,\text{ and }c\text{ are real numbers and }a\neq 0\nonumber\]
To solve quadratic equations we need methods different from the ones we used in solving linear equations. We will look at one method here and then several others in a later chapter.
Use the Zero Product Property
We will first solve some quadratic equations by using the Zero Product Property . The Zero Product Property says that if the product of two quantities is zero, then at least one of the quantities is zero. The only way to get a product equal to zero is to multiply by zero itself.
ZERO PRODUCT PROPERTY
If \(a·b=0\), then either \(a=0\) or \(b=0\) or both.
We will now use the Zero Product Property, to solve a quadratic equation .
Example \(\PageIndex{1}\): How to Solve a Quadratic Equation Using the Zero Product Property
Solve: \((5n−2)(6n−1)=0\).
- Answer
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Example \(\PageIndex{2}\)
Solve: \((3m−2)(2m+1)=0\).
- Answer
-
\(m=\frac{2}{3},\space m=−\frac{1}{2}\)
Example \(\PageIndex{3}\)
Solve: \((4p+3)(4p−3)=0\).
- Answer
-
\(p=−\frac{3}{4},\space p=\frac{3}{4}\)
USE THE ZERO PRODUCT PROPERTY.
- Set each factor equal to zero.
- Solve the linear equations.
- Check.
Solve Quadratic Equations by Factoring
The Zero Product Property works very nicely to solve quadratic equations. The quadratic equation must be factored, with zero isolated on one side. So we be sure to start with the quadratic equation in standard form , \(ax^2+bx+c=0\). Then we factor the expression on the left.
Solve: \(2y^2=13y+45\).
- Answer
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Example \(\PageIndex{5}\)
Solve: \(3c^2=10c−8\).
- Answer
-
\(c=2,\space c=\frac{4}{3}\)
Example \(\PageIndex{6}\)
Solve: \(2d^2−5d=3\).
- Answer
-
\(d=3,\space d=−12\)
SOLVE A QUADRATIC EQUATION BY FACTORING.
- Write the quadratic equation in standard form, \(ax^2+bx+c=0\).
- Factor the quadratic expression.
- Use the Zero Product Property.
- Solve the linear equations.
- Check. Substitute each solution separately into the original equation.
Before we factor, we must make sure the quadratic equation is in standard form .
Solving quadratic equations by factoring will make use of all the factoring techniques you have learned in this chapter! Do you recognize the special product pattern in the next example?
Example \(\PageIndex{7}\)
Solve: \(169q^2=49\).
- Answer
-
\(\begin{array} {ll} &169x^2=49 \\ \text{Write the quadratic equation in standard form.} &169x^2−49=0 \\ \text{Factor. It is a difference of squares.} &(13x−7)(13x+7)=0 \\ \text{Use the Zero Product Property to set each factor to }0. & \\ \text{Solve each equation.} &\begin{array} {ll} 13x−7=0 &13x+7=0 \\ 13x=7 &13x=−7 \\ x=\frac{7}{13} &x=−\frac{7}{13} \end{array} \end{array}\)
Check:
We leave the check up to you.
Example \(\PageIndex{8}\)
Solve: \(25p^2=49\).
- Answer
-
\(p=\frac{7}{5},p=−\frac{7}{5}\)
Example \(\PageIndex{9}\)
Solve: \(36x^2=121\).
- Answer
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\(x=\frac{11}{6},x=−\frac{11}{6}\)
In the next example, the left side of the equation is factored, but the right side is not zero. In order to use the Zero Product Property , one side of the equation must be zero. We’ll multiply the factors and then write the equation in standard form.
Example \(\PageIndex{10}\)
Solve: \((3x−8)(x−1)=3x\).
- Answer
-
\(\begin{array} {ll} &(3x−8)(x−1)=3x \\ \text{Multiply the binomials.} &3x^2−11x+8=3x \\ \text{Write the quadratic equation in standard form.} &3x^2−14x+8=0 \\ \text{Factor the trinomial.} &(3x−2)(x−4)=0 \\ \begin{array} {l} \text{Use the Zero Product Property to set each factor to 0.} \\ \text{Solve each equation.} \end{array} &\begin{array} {ll} 3x−2=0 &x−4=0 \\ 3x=2 &x=4 \\ x=\frac{2}{3} & \end{array} \\ \text{Check your answers.} &\text{The check is left to you.} \end{array}\)
Example \(\PageIndex{11}\)
Solve: \((2m+1)(m+3)=12m\).
- Answer
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\(m=1,\space m=\frac{3}{2}\)
Example \(\PageIndex{12}\)
Solve: \((k+1)(k−1)=8\).
- Answer
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\(k=3,\space k=−3\)
In the next example, when we factor the quadratic equation we will get three factors. However the first factor is a constant. We know that factor cannot equal 0.
Example \(\PageIndex{13}\)
Solve: \(3x^2=12x+63\).
- Answer
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\(\begin{array} {ll} &3x^2=12x+63 \\ \text{Write the quadratic equation in standard form.} &3x^2−12x−63=0 \\ \text{Factor the greatest common factor first.} &3(x^2−4x−21)=0 \\ \text{Factor the trinomial.} &3(x−7)(x+3)=0 \\ \begin{array} {l} \text{Use the Zero Product Property to set each factor to 0.} \\ \text{Solve each equation.} \end{array} &\begin{array} {lll} 3\neq 0 &x−7=0 &x+3=0 \\ 3\neq 0 &x=7 &x=−3 \end{array} \\ \text{Check your answers.} &\text{The check is left to you.} \end{array}\)
Example \(\PageIndex{14}\)
Solve: \(18a^2−30=−33a\).
- Answer
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\(a=−\frac{5}{2},a=\frac{2}{3}\)
Example \(\PageIndex{15}\)
Solve: \(123b=−6−60b^2\)
- Answer
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\(b=−2,\space b=−\frac{1}{20}\)
The Zero Product Property also applies to the product of three or more factors. If the product is zero, at least one of the factors must be zero. We can solve some equations of degree greater than two by using the Zero Product Property, just like we solved quadratic equations.
Example \(\PageIndex{16}\)
Solve: \(9m^3+100m=60m^2\)
- Answer
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\(\begin{array} {ll} & 9m^3+100m=60m^2 \\ \text{Bring all the terms to one side so that the other side is zero.} &9m^3−60m^2+100m=0 \\ \text{Factor the greatest common factor first.} &m(9m^2−60m+100)=0 \\ \text{Factor the trinomial.} &m(3m−10)^2=0 \end{array}\\ \begin{array} {l} \text{Use the Zero Product Property to set each factor to 0.} \\ \text{Solve each equation.} &\begin{array} {lll} m=0 &3m−10=0 &{}\\ m=0 &m=\frac{10}{3} & {} \end{array}\\ \text{Check your answers.} &\text{The check is left to you.} \end{array}\)
Example \(\PageIndex{17}\)
Solve: \(8x^3=24x^2−18x\).
- Answer
-
\(x=0,\space x=\frac{3}{2}\)
Example \(\PageIndex{18}\)
Solve: \(16y^2=32y^3+2y\).
- Answer
-
\(y=0,\space y=14\)
Solve Equations with Polynomial Functions
As our study of polynomial functions continues, it will often be important to know when the function will have a certain value or what points lie on the graph of the function. Our work with the Zero Product Property will be help us find these answers.
Example \(\PageIndex{19}\)
For the function \(f(x)=x^2+2x−2\),
ⓐ find \(x\) when \(f(x)=6\)
ⓑ find two points that lie on the graph of the function.
- Answer
-
ⓐ
\(\begin{array} {ll} &f(x)=x^2+2x−2 \\ \text{Substitute }6\text{ for }f(x). &6=x^2+2x−2 \\ \text{Put the quadratic in standard form.} &x^2+2x−8=0 \\ \text{Factor the trinomial.} &(x+4)(x−2)=0 \\ \begin{array} {l} \text{Use the zero product property.} \\ \text{Solve.} \end{array} &\begin{array} {lll} x+4=0 &\text{or} &x−2=0 \\ x=−4 &\text{or} &x=2 \end{array} \\ \text{Check:} & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ \begin{array} {lll} \quad &\hspace{3mm} f(x)=x^2+2x−2 &f(x)=x^2+2x−2 \\ \quad &f(−4)=(−4)^2+2(−4)−2 &f(2)=2^2+2·2−2 \\ \quad &f(−4)=16−8−2 &f(2)=4+4−2 \\ \quad &f(−4)=6\checkmark &f(2)=6\checkmark \end{array} & \end{array} \)
ⓑ Since \(f(−4)=6\) and \(f(2)=6\), the points \((−4,6)\) and \((2,6)\) lie on the graph of the function.
Example \(\PageIndex{20}\)
For the function \(f(x)=x^2−2x−8\),
ⓐ find \(x\) when \(f(x)=7\)
ⓑ Find two points that lie on the graph of the function.
- Answer
-
ⓐ \(x=−3\) or \(x=5\)
ⓑ \((−3,7)\space (5,7)\)
Example \(\PageIndex{21}\)
For the function \(f(x)=x^2−8x+3\),
ⓐ find \(x\) when \(f(x)=−4\)
ⓑ Find two points that lie on the graph of the function.
- Answer
-
ⓐ \(x=1\) or \(x=7\)
ⓑ \((1,−4)\space (7,−4)\)
The Zero Product Property also helps us determine where the function is zero. A value of \(x\) where the function is \(0\), is called a zero of the function .
ZERO OF A FUNCTION
For any function \(f\), if \(f(x)=0\), then \(x\) is a zero of the function .
When \(f(x)=0\), the point \((x,0)\) is a point on the graph. This point is an \(x\)- intercept of the graph. It is often important to know where the graph of a function crosses the axes. We will see some examples later.
Example \(\PageIndex{22}\)
For the function \(f(x)=3x^2+10x−8\), find
ⓐ the zeros of the function,
ⓑ any \(x\)-intercepts of the graph of the function
ⓒ any \(y\)-intercepts of the graph of the function
- Answer
-
ⓐ To find the zeros of the function, we need to find when the function value is 0.
\(\begin{array} {ll} &f(x)=3x^2+10x−8 \\ \text{Substitute }0\text{ for}f(x). &0=3x^2+10x−8 \\ \text{Factor the trinomial.} &(x+4)(3x−2)=0 \\ \begin{array} {l} \text{Use the zero product property.} \\ \text{Solve.} \end{array} &\begin{array} {lll} x+4=0 &\text{or} &3x−2=0 \\ x=−4 &\text{or} &x=\frac{2}{3} \end{array} \end{array}\)
ⓑ An \(x\)-intercept occurs when \(y=0\). Since \(f(−4)=0\) and \(f(\frac{2}{3})=0\), the points \((−4,0)\) and \((\frac{2}{3},0)\) lie on the graph. These points are \(x\)-intercepts of the function.
ⓒ A \(y\)-intercept occurs when \(x=0\). To find the \(y\)-intercepts we need to find \(f(0)\).
\(\begin{array} {ll} &f(x)=3x^2+10x−8 \\ \text{Find }f(0)\text{ by substituting }0\text{ for }x. &f(0)=3·0^2+10·0−8 \\ \text{Simplify.} &f(0)=−8 \end{array} \)
Since \(f(0)=−8\), the point \((0,−8)\) lies on the graph. This point is the \(y\)-intercept of the function.
Example \(\PageIndex{23}\)
For the function \(f(x)=2x^2−7x+5\), find
ⓐ the zeros of the function
ⓑ any \(x\)-intercepts of the graph of the function
ⓒ any \(y\)-intercepts of the graph of the function.
- Answer
-
ⓐ \(x=1\) or \(x=\frac{5}{2}\)
ⓑ \((1,0),\space (\frac{5}{2},0)\) ⓒ \((0,5)\)
Example \(\PageIndex{24}\)
For the function \(f(x)=6x^2+13x−15\), find
ⓐ the zeros of the function
ⓑ any \(x\)-intercepts of the graph of the function
ⓒ any \(y\)-intercepts of the graph of the function.
- Answer
-
ⓐ \(x=−3\) or \(x=\frac{5}{6}\)
ⓑ \((−3,0),\space (\frac{5}{6},0)\) ⓒ \((0,−15)\)
Solve Applications Modeled by Polynomial Equations
The problem-solving strategy we used earlier for applications that translate to linear equations will work just as well for applications that translate to polynomial equations. We will copy the problem-solving strategy here so we can use it for reference.
USE A PROBLEM SOLVING STRATEGY TO SOLVE WORD PROBLEMS.
- Read the problem. Make sure all the words and ideas are understood.
- Identify what we are looking for.
- Name what we are looking for. Choose a variable to represent that quantity.
- Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation.
- Solve the equation using appropriate algebra techniques.
- Check the answer in the problem and make sure it makes sense.
- Answer the question with a complete sentence.
We will start with a number problem to get practice translating words into a polynomial equation.
Example \(\PageIndex{25}\)
The product of two consecutive odd integers is 323. Find the integers.
- Answer
-
\(\begin{array} {ll} \textbf{Step 1. Read }\text{the problem.} & \\ \textbf{Step 2. Identify }\text{what we are looking for.} &\text{We are looking for two consecutive integers.} \\ \textbf{Step 3. Name}\text{ what we are looking for.} &\text{Let } n=\text{ the first integer.} \\ &n+2= \text{ next consecutive odd integer} \\ \begin{array} {l} \textbf{Step 4. Translate }\text{into an equation. Restate the}\hspace{20mm} \\ \text{problem in a sentence.} \end{array} &\begin{array} {l} \text{The product of the two consecutive odd} \\ \text{integers is }323. \end{array} \\ &\quad n(n+2)=323 \\ \textbf{Step 5. Solve }\text{the equation.} n^2+2n=323 \\ \text{Bring all the terms to one side.} &n^2+2n−323=0 \\ \text{Factor the trinomial.} &(n−17)(n+19)=0 \\ \begin{array} {l} \text{Use the Zero Product Property.} \\ \text{Solve the equations.} \end{array} &\begin{array} {ll} n−17=0 \hspace{10mm}&n+19=0 \\ n=17 &n=−19 \end{array} \end{array} \)
There are two values for \(n\) that are solutions to this problem. So there are two sets of consecutive odd integers that will work.\(\begin{array} {ll} \text{If the first integer is } n=17 \hspace{60mm} &\text{If the first integer is } n=-19 \\ \text{then the next odd integer is} &\text{then the next odd integer is} \\ \hspace{53mm} n+2 &\hspace{53mm} n+2 \\ \hspace{51mm} 17+2 &\hspace{51mm} -19+2 \\ \hspace{55mm} 19 &\hspace{55mm} -17 \\ \hspace{51mm} 17,19 &\hspace{51mm} -17,-19 \\ \textbf{Step 6. Check }\text{the answer.} & \\ \text{The results are consecutive odd integers} & \\ \begin{array} {ll} 17,\space 19\text{ and }−19,\space −17. & \\ 17·19=323\checkmark &−19(−17)=323\checkmark \end{array} & \\ \text{Both pairs of consecutive integers are solutions.} & \\ \textbf{Step 7. Answer }\text{the question} &\text{The consecutive integers are }17, 19\text{ and }−19,−17. \end{array} \)
Example \(\PageIndex{26}\)
The product of two consecutive odd integers is 255. Find the integers.
- Answer
-
\(−15,−17\) and \(15, 17\)
Example \(\PageIndex{27}\)
The product of two consecutive odd integers is 483 Find the integers.
- Answer
-
\(−23,−21\) and \(21, 23\)
Were you surprised by the pair of negative integers that is one of the solutions to the previous example? The product of the two positive integers and the product of the two negative integers both give positive results.
In some applications, negative solutions will result from the algebra, but will not be realistic for the situation.
Example \(\PageIndex{28}\)
A rectangular bedroom has an area 117 square feet. The length of the bedroom is four feet more than the width. Find the length and width of the bedroom.
- Answer
-
Step 1. Read the problem. In problems involving
geometric figures, a sketch can help you visualize
the situation.Step 2. Identify what you are looking for. We are looking for the length and width. Step 3. Name what you are looking for. Let \(w=\text{ the width of the bedroom}\). The length is four feet more than the width. \(w+4=\text{ the length of the garden}\) Step 4. Translate into an equation. Restate the important information in a sentence. The area of the bedroom is 117 square feet. Use the formula for the area of a rectangle. \(A=l·w\) Substitute in the variables. \(117=(w+4)w\) Step 5. Solve the equation Distribute first. \(117=w^2+4w\) Get zero on one side. \(117=w^2+4w\) Factor the trinomial. \(0=w^2+4w−117\) Use the Zero Product Property. \(0=(w^2+13)(w−9)\) Solve each equation. \(0=w+13\quad 0=w−9\) Since \(w\) is the width of the bedroom, it does not
make sense for it to be negative. We eliminate that value for \(w\).\(\cancel{w=−13}\) \(\quad w=9\) \(w=9\) The width is 9 feet. Find the value of the length. \(w+4\)
\(9+4\)
13 The length is 13 feet.Step 6. Check the answer.
Does the answer make sense?
Yes, this makes sense.Step 7. Answer the question. The width of the bedroom is 9 feet and
the length is 13 feet.
Example \(\PageIndex{29}\)
A rectangular sign has area 30 square feet. The length of the sign is one foot more than the width. Find the length and width of the sign.
- Answer
-
The width is 5 feet and length is 6 feet.
Example \(\PageIndex{30}\)
A rectangular patio has area 180 square feet. The width of the patio is three feet less than the length. Find the length and width of the patio.
- Answer
-
The length of the patio is 12 feet and the width 15 feet.
In the next example, we will use the Pythagorean Theorem \((a^2+b^2=c^2)\). This formula gives the relation between the legs and the hypotenuse of a right triangle.
We will use this formula to in the next example.
Example \(\PageIndex{31}\)
A boat’s sail is in the shape of a right triangle as shown. The hypotenuse will be 17 feet long. The length of one side will be 7 feet less than the length of the other side. Find the lengths of the sides of the sail.
- Answer
-
Step 1. Read the problem Step 2. Identify what you are looking for. We are looking for the lengths of the
sides of the sail.Step 3. Name what you are looking for.
One side is 7 less than the other.Let \(x=\text{ length of a side of the sail}\).
\(x−7=\text{ length of other side}\)Step 4. Translate into an equation. Since this is a
right triangle we can use the Pythagorean Theorem.\(a^2+b^2=c^2\) Substitute in the variables. \(x^2+(x−7)^2=17^2\) Step 5. Solve the equation
Simplify.\(x^2+x^2−14x+49=289\) \(2x^2−14x+49=289\) It is a quadratic equation, so get zero on one side. \(2x^2−14x−240=0\) Factor the greatest common factor. \(2(x^2−7x−120)=0\) Factor the trinomial. \(2(x−15)(x+8)=0\) Use the Zero Product Property. \(2\neq 0\quad x−15=0\quad x+8=0\) Solve. \(2\neq 0\quad x=15\quad x=−8\) Since \(x\) is a side of the triangle, \(x=−8\) does not
make sense.\(2\neq 0\quad x=15\quad \cancel{x=−8}\) Find the length of the other side. If the length of one side is
then the length of the other side is
8 is the length of the other side.Step 6. Check the answer in the problem
Do these numbers make sense?
Step 7. Answer the question The sides of the sail are 8, 15 and 17 feet.
Example \(\PageIndex{32}\)
Justine wants to put a deck in the corner of her backyard in the shape of a right triangle. The length of one side of the deck is 7 feet more than the other side. The hypotenuse is 13. Find the lengths of the two sides of the deck.
- Answer
-
5 feet and 12 feet
Example \(\PageIndex{33}\)
A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of the other leg. Find the lengths of the hypotenuse and the other leg.
- Answer
-
24 feet and 25 feet
The next example uses the function that gives the height of an object as a function of time when it is thrown from 80 feet above the ground.
Example \(\PageIndex{34}\)
Dennis is going to throw his rubber band ball upward from the top of a campus building. When he throws the rubber band ball from 80 feet above the ground, the function \(h(t)=−16t^2+64t+80\) models the height, \(h\), of the ball above the ground as a function of time, \(t\). Find:
ⓐ the zeros of this function which tell us when the ball hits the ground
ⓑ when the ball will be 80 feet above the ground
ⓒ the height of the ball at \(t=2\) seconds.
- Answer
-
ⓐ The zeros of this function are found by solving \(h(t)=0\). This will tell us when the ball will hit the ground.
\(\begin{array} {ll} &h(t)=0 \\ \text{Substitute in the polynomial for }h(t). &−16t^2+64t+80=0 \\ \text{Factor the GCF, }−16. &−16(t^2−4t−5)=0 \\ \text{Factor the trinomial.} &−16(t−5)(t+1)=0 \\ \begin{array} {l} \text{Use the Zero Product Property.} \\ \text{Solve.} \end{array} &\begin{array} {ll} t−5=0 &t+1=0 \\ t=5 &t=−1 \end{array} \end{array} \)The result \(t=5\) tells us the ball will hit the ground 5 seconds after it is thrown. Since time cannot be negative, the result \(t=−1\) is discarded.
ⓑ The ball will be 80 feet above the ground when \(h(t)=80\).
\(\begin{array} {ll} &h(t)=80 \\ \text{Substitute in the polynomial for }h(t). &−16t^2+64t+80=80 \\ \text{Subtract 80 from both sides.} &−16t^2+64t=0 \\ \text{Factor the GCF, }−16t. &−16t(t−4)=0 \\ \begin{array} {l} \text{Use the Zero Product Property.} \\ \text{Solve.}\end{array} &\begin{array} {ll} −16t=0 &t−4=0 \\ t=0 &t=4 \end{array} \\ &\text{The ball will be at 80 feet the moment Dennis} \\ &\text{tosses the ball and then 4 seconds later, when} \\ &\text{the ball is falling.} \end{array} \)ⓒ To find the height ball at \(t=2\) seconds we find \(h(2)\).
\(\begin{array} {ll} &h(t)=−16t^2+64t+80 \\ \text{To find }h(2)\text{ substitute }2\text{ for }t. &h(2)=−16(2)^2+64·2+80 \\ \text{Simplify.} &h(2)=144 \\ &\text{After 2 seconds, the ball will be at 144 feet.} \end{array}\)
Example \(\PageIndex{35}\)
Genevieve is going to throw a rock from the top a trail overlooking the ocean. When she throws the rock upward from 160 feet above the ocean, the function \(h(t)=−16t^2+48t+160\) models the height, \(h\), of the rock above the ocean as a function of time, \(t\). Find:
ⓐ the zeros of this function which tell us when the rock will hit the ocean
ⓑ when the rock will be 160 feet above the ocean.
ⓒ the height of the rock at \(t=1.5\) seconds.
- Answer
-
ⓐ 5 ⓑ 0;3 ⓒ 196
Example \(\PageIndex{36}\)
Calib is going to throw his lucky penny from his balcony on a cruise ship. When he throws the penny upward from 128 feet above the ground, the function \(h(t)=−16t^2+32t+128\) models the height, \(h\), of the penny above the ocean as a function of time, \(t\). Find:
ⓐ the zeros of this function which is when the penny will hit the ocean
ⓑ when the penny will be 128 feet above the ocean.
ⓒ the height the penny will be at \(t=1\) seconds which is when the penny will be at its highest point.
- Answer
-
ⓐ 4 ⓑ 0;2 ⓒ 144
Access this online resource for additional instruction and practice with quadratic equations.
Key Concepts
- Polynomial Equation: A polynomial equation is an equation that contains a polynomial expression. The degree of the polynomial equation is the degree of the polynomial.
-
Quadratic Equation:
An equation of the form \(ax^2+bx+c=0\) is called a quadratic equation.
\[a,b,c\text{ are real numbers and } a\neq 0\nonumber\]
- Zero Product Property: If \(a·b=0\), then either \(a=0\) or \(b=0\) or both.
-
How to use the Zero Product Property
- Set each factor equal to zero.
- Solve the linear equations.
- Check.
-
How to solve a quadratic equation by factoring.
- Write the quadratic equation in standard form, \(ax^2+bx+c=0\).
- Factor the quadratic expression.
- Use the Zero Product Property.
- Solve the linear equations.
- Check. Substitute each solution separately into the original equation.
- Zero of a Function: For any function \(f\), if \(f(x)=0\), then \(x\) is a zero of the function.
-
How to use a problem solving strategy to solve word problems.
- Read the problem. Make sure all the words and ideas are understood.
- Identify what we are looking for.
- Name what we are looking for. Choose a variable to represent that quantity.
- Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation.
- Solve the equation using appropriate algebra techniques.
- Check the answer in the problem and make sure it makes sense.
- Answer the question with a complete sentence.