Unit 5: Quadratic Equations
Unit 5: Quadratic Equations
Solutions to Quadratic Equations
Objectives:
E.1To solve quadratic equations using the quadratic formula.
E.2To solve quadratic equations having complex roots.
E.3To solve word problems involving real world applications of quadratic equations.
Notes:
We may recall from earlier grades that a QuadraticEquation is an equation of the general form; ax2 +bx + c = 0, Where: a, b, and c are real numbers.
The solutions or roots of the equation can be found in several ways;
1.Factoring:
For Example: Solve:
(2x - 3)(x + 4) = 0Factored
2x - 3 = 0, or x + 4 = 0Set each factor = 0.
2x = 3, x = 3/2Solve each factor for x.
or x + 4 = 0, x = -4
x = 3/2 or x = -4The solution.
2.Completing the square:
For Example: Solve: w
/ Divide each term by “a”/ Move the constant term across the = sign
and add to both sides.
/ Factor the left hand side.
/ Simplify the right hand side.
/ Square root both sides.
/ Solve for x
3.The Quadratic Formula:
Derived by completing the square on the general quadratic formula.
For Example: Solve 3x2 - x - 10 = 0
/ Write The Quadratic Equation/ Substitute valuers for a, b, and c
/ Simplify.
or / Solve.
We may also remember that the roots to a quadratic equation are represented by the x –intercepts of the related function:
For Example: The quadratic
Written as a relation of x & y.Table of values
x
-4
-2
0
2
4
6
8
y
15
5
-1
-3
-1
5
15
x- intercepts from graph;
x = -0.5, x = 4.5
Notice that the values are estimates, at best. /
TI-82 Calculator: Remember that we can graph equations easily using the [Y=] key.
To graph the equation,
-Press [Y=], use [CLEAR] to remove any functions already entered
-Enter (1/2)X2-2X-1 into Y1. Use the [X,T,] key to enter the variable X.
-To see the graph, press [GRAPH].
-If the axes do not allow you to see the graph, press [ZOOM] and choose 6:ZStandard to reset the axes.
Solving Problems
Without doubt, the Quadratic Formula is one of the most useful formulae developed in high school mathematics. This formula allows us to consistently and (relatively) easily solve any quadratic equation. Although the examples given in algebra text books may appear contrived, quadratic equations and their solutions are a part of any activity in which algebra is used.
For Example: The distance traveled by an accelerating body is given in the formula;
If an anti aircraft gun fires upward with a muzzle velocity of 500 m/s, how long will it take the shell to reach an aircraft flying at 1500 m height?
a = acceleration due to gravity = - 9.81 m/s2
Solution: Given: d = 1500 m, vo = 500 m/s, and a = -9.81 m/s2
Find: time = t
/ Substituting into :/ In standard quadratic form.
/ The quadratic formula.
/ Substituting into the quadratic formula.
/ Simplify.
/ Solved for t.
There are two answers, t = 3.1 s or t = 98.9 s. 3.1 s is the time for the shell to reach 1500 m on the way up, 98.9 s is the time that the shell arrives on the way down if it misses the plane. Something that an artillery officer had better be aware of before firing!
The quadratic equation will produce two answers for many questions. You must be sure to check which answer is the correct one by asking “Is this answer possible or appropriate?” for both.
It is also a useful exercise to try to determine the meaning of an excluded answer. In the example given, the excluded answer provides warning that the artillery shell may hit some undesired target on the way down.
Always check your answer to see if it is reasonable.
TI-82 Calculator: To key in the quadratic formula, one needs to use brackets carefully.
To solve the quadratic , one should enter the following;Remember that the square root is: [2nd]-[x2]
(2+(22-4*.5*-1))/(2*.5) [ENTER] ,
If keyed in correctly the result should be: 4.449489743
To find the other root, use [2nd]-[ENTER] to recall the last entry and use the arrow keys to change the sign;
(2- (22-4*.5*-1))/(2*.5) [ENTER] ,
If keyed in correctly the result should be: -.4494897428
**It is also possible to us the [TRACE] function or the CALC menu ([2nd]-[TRACE]) to find the roots to a quadratic function that has been graphed.
Exact vs Approximate Roots
When exact roots are specified, the answer must be left in radical form:
For Example: Solve the quadratic , leave the roots in exact form.
Solution: / Write the quadratic formula first/ Substitute in values for a, b, & c
/ Simplify
/ Leave all radicals in reduced form.
Approximate Roots are found by using a calculator to find the decimal value of the solution. This is the form required by most applications or problems.
If is the exact solution then
x = 2 + 2.449 = 4.449, or
x = 2 - 2.449 = -0.449 is the approximate solution.
Always round your answer to three decimal places or less. Many questions will specify the required precision of the answer.
Complex Roots
When using the quadratic equation, the result may sometimes include a negative square root. In this case, the roots are Non-real or Complex roots. They should be expressed in reduced form as x = a + bi.
For Example: Solve 3x2 - 4x + 2 = 0
-1-
Unit 5: Quadratic Equations
Solution: / Write the quadratic formula first/ Substitute in values for a, b, & c
/ Simplify
/ A negative square root
/ Express in terms of "i"
/ Simplify the root.
/ Remove any common factors.
/ Express in standard form.
Practice Questions 1: Using the Quadratic Formula
1.Solve each of the equations below using the quadratic formula. Leave the roots in exact form.
a. x2 - 8x + 16 = 0b. 2x2 + x - 5 = 0
c. q2- 5x - 1= 0d. 3s2 + 7x = -4
e. x2 + 64 = 0f. 27 + r2 = 0
g. 3t2 - 2t + 3 = 0h. 4m2 = 6m +3
i.j. -2(x+1)2 = -2
k. 0.1a2 + 0.14a - 23 = 0 (decimal answer)l.
m.n.
2.Sketch the graph of each of the following and estimate the roots.
a. x2 - 6x + 9 = yb. 6x2 + 11x = 10 + y
c. 3x2 - 2x + 1 = yd. -3x2 - 4x + 2 = y
3.Use the quadratic formula to find the exact roots for each of the questions in 2.
4.From your answers to questions 2 and 3, can you make a general statement about the relationship between the graph of a function and the number and nature of roots?
Solving Problems Involving Quadratics
As mentioned earlier, quadratics are a part of many different fields of study. To solve problems we should;
1.Read the problem to identify what we are being given and what we are asked to find.
2.Use a “Let” statement and/or a diagram to identify the variable(s).
3.Write a numerical statement that describes the problem.
4. Solve: In the case of a quadratic we can solve by:
-simplifying the equation
-put the equation into standard form: ax2 + bx + c = 0
-substitute into the quadratic formula:
-solve to find both roots.
-check to determine which roots are extraneous (not applicable)
5.Check, is the answer reasonable? Does it work?
6.Write the answer in a sentence.
For Example: The hypotenuse of a right triangle is 24 m long. If one leg is 8 m shorter than the other find the length of both legs.
Solution: Let x = the length of the longest legx - 8 = the shorter leg.
From Pythagoras;
242 = x2 + (x - 8)2 / Diagram
-1-
Unit 5: Quadratic Equations
576 = x2 + x2 - 16x + 64Expand and Simplify
0 = 2x2 - 16x - 512Put in standard form.
Write the quadratic formula.
Substitute into the formula.
Simplify
The exact answer.
x = 20.83 m or x =mThe decimal answer.
x = 20.83, x - 8 = 12. 83The solution.
242 = 20.832+ 12.8323Check
The legs of the triangle are 20.83 m and 12. 83 m.
For Example: The winning pair of drivers in an automobile rally from Victoria to Winnipeg took 24 hours, 30 minutes to complete the trip. The first driver drove 900 km through the mountains, while the second driver drove the final 1500 km at a rate of speed 50 km/h less then twice the rate of the first driver. What was the average rate of speed for each driver?
Solution: Let x = the speed of the first driver.
Then 2x - 50 = the speed of the second driver.
If , then, and time for driver 1 + time for driver 2 = 24.5 h.
24.5 h =The Equation.
Multiplying each term by x(2x - 50).
Simplify
In standard form.
The Quadratic formula.
Substitute into the formula.
Simplify.
x = 81.01 or x = 11.34The decimal answer.
If x = 81.01, then 2x - 50 = 112.02
If x = 11.34, then 2x - 50 = -27.32Discard this solution.
h3Check.
The first driver averaged 81.01 km/h through the mountains and the second averaged 112.02km/h for the rest of the trip.
Some Useful Formulae
Pythagoras: Area of a triangle:
Speed: , , Area of a circle:,
Area of a rectangle: Area of a sphere:
Area of a cylinder:
Practice Questions 2: Word Problems
Give both the exact and decimal values of the solutions.
1.The sum of the squares of two consecutive odd integers is 650. Find the numbers.
(Consecutive odd integers are x, and x + 2)
2.The length of a blanket is 0.5 m greater than its width. If the blanket has an area of 3 m2, find the dimensions of the blanket.
3.A cylindrical drum’s radius is 0.10 m less than its height. If the total area of the drum is 4.28m2, find the radius and height of the drum.
(, give a decimal answer to 3 decimal places)
4.The diagonal of a square is 6 centimeters longer than its sides. Find the dimensions of the square.
5.One number is five greater than another. The product of the numbers is 59 greater than ten times their average. Find the numbers.
6.A rectangular yard has an area of 560 m2 and a perimeter of 96 m. What are the dimensions of the lot?
7.A cultural interpretation center has two teepees on display. The radius of one teepee is 1.24 meters larger than the other. If the area within the larger teepee is exactly double the smaller one, what is the radius of the teepees.( = 3.14, answer to 2 decimal places) /
8.A painter has a 3 m ladder. He has placed it against the house so that the distance from the bottom of the ladder to the house is exactly half the distance that the ladder extends up the wall. How far does the ladder reach up the wall? /
9.A cyclist has taken a 180 km trip. She traveled the first 120 km at one rate and finished the trip at a speed 15.2 km/h slower. If the trip took 3 hours, what was her average speed on each part?
10.A canoeist paddles 20 km up stream on a river and the same distance back. The current flows at 3km/h. If the entire trip took 7 hours, what is the speed of the canoeist in still water?
11.Extra for experts. Derive the quadratic formula by completing the square on the general quadratic, ax2 + bx + c = 0.
Objectives:
E.4To determine the nature of the roots of a quadratic equation using the discriminant.
E.5To determine the sum and the product of the roots of a quadratic equation.
E.6To write a quadratic equation, given the roots.
E.7To solve equations of degree other than two by expressing them in quadratic form.
E.8To solve quadratic inequalities.
Notes:
The Discriminant
From the equations solved so far in this section, we are able to form a general idea of the kinds of roots that a quadratic might have;
Equation:Quadratic FormulaRoots
i.
ii., or
iii., or
iv., or
The four examples above show how the nature and number of the roots can be found. The key is the value produced by or, to be more precise, the value of b2-4ac . This is called the Discriminant (D) of the quadratic.
If D = 0, then there is only one real root.
If D > 0, and D is a perfect square, then there are two rational roots.
If D > 0, and D is not a perfect square, then there are two irrational roots.
If D < 0, then there are no real roots. (two complex roots)
For Example: What is the nature of the roots of the quadratic 3x2 - 12x + 12 = 0?
Solution: Find the value of the discriminantD = b2 - 4ac = 122 - 4(3)(12) = 0
There is one real root.
For Example: Find the values of k for which the roots of 2x2 + kx + 6 = 0 are complex.
Solution: Use the fact that for D < 0, the roots are not real.
b2 - 4ac < 0 k2 - 4(2)(6) < 0Substituting into the formula.
k2 - 48 < 0Simplifying
k2 < 48
k < or k >Remember that a square root has two values.
k < or k >Reduce all radicals to lowest form.
The Sum and Product of the Roots
For the equation x2 - 4x - 21 = 0, the roots are x = 7 or i = -3.
The sum of the roots is 7 + (-3) = 4Notice that this is -(-4) or -b from the equation
The product of the roots is (7)(-3) = -21Notice that this is the same as c in the equation.
For the equation 2x2 + x - 15 = 0, the roots are; x = and, x = -3.
The sum of the roots is - 3 =The same as from the equation.
The product of the roots is •(-3) =The same as from the equation.
In general, for any quadratic ax2 + bx + c = 0, with roots x = r1 and x = r2:
The sum of the roots is: and
The product of the roots is:
For Example: Find the roots for the equation 3x2 - 2x + 5 = 0. Show that the sum of the roots is and that the product of the roots is.
Solution: Use the quadratic formula to find the roots:
The quadratic formula.
Substituting into the formula.
Simplify (Note that D = -56 complex roots)
Simplified and expressed as a complex number
**The sum of the roots =
**Simplified, the product of the roots =
Writing The Equation of a Quadratic
We can use the sum and product of the roots of a quadratic to write the corresponding equation. If the roots are x = r1 and x = r2, then the equation is;
For Example: Write the equation of the quadratic having the rootsand .
Solution: / Place the sum and product of the roots into the quadratic equation./ Calculate the sum and product.
/ Clear the fractions by multiplying by the denominator.
For Example: Write the equation of the quadratic having the roots
Solution: / Place the sum and product of the roots into the quadratic equation./ Calculate the sum and product.
For Example: Write the equation of the quadratic having the roots
Solution: / Place the sum and product of the roots into the quadratic equation./ Calculate the sum and product.
/ Clear the fractions by multiplying by the denominator.
Notice that, in most cases, the sum and product of the roots will work out not to have any radicals or imaginary components.
In every case it is customary to clear all fractions from the final equation. This is achieved by multiplying all terms by the least common denominator (L.C.D.).
Practice Questions 3: The Determinant and the roots of a Quadratic.
1.Use the value of the discriminant to determine the nature of the roots for each of the following quadratics;
a. 2x2 - 3x + 5 = 0b. 2x2 - 3x - 5 = 0
c. -4x2 + 8x = 4d.
e. -2x2 - 3x - 1f. 2ix2 - 5x = -3i
2.Given the formula for the general quadratic is ax2 + bx + c = 0, determine the nature of the roots if;
a. a is negative and c is positiveb. c is negative and a is positive
c. a and c are positive and b =
3.For what value of k are the roots of:
a. realb. complex
c. equal
4. For what values of s are the roots of:
a. realb. complex
c. equal
5.Predict the sum and product of the roots for each of the following. Check by finding the roots.
a. x2 - 3x - 28 = 0b. x2 + 5x = 24
c. 3x2 - 8x + 5 = 0d.
e.f. 5x2 + 3x = -7
6.a. add:
b. multiply;
c. comment.
7.Write the equation of a quadratic having the following roots:
a. {3, -2}b. {7, 11}
c.d.
e.f.
g. {3k, -2m}h.
Solving Equations of Degree Other Than 2
The techniques used in solving quadratics can be used to solve any equation having the same pattern of terms as a quadratic.
For Example: Solve x3- 5x2 + 6x = 0
Solution: Always start by looking for common factors.
x(x2 - 5x + 6) = 0Factoring to remove the common factor, (x).
x(x - 2)(x - 3) = 0Factoring the quadratic.
x = 0, or x - 2 = 0, or x - 3 = 0Using the property that if ab = 0, a = 0 or b = 0
x = 0, or x = 2 or x = 3Solve for x.
Once the common factor is removed, the quadratic can be solved using any of the methods including the quadratic formula.
For Example: Solve x4 - 13x2 + 36 = 0
Solution: Exponents that are multiples of the exponents in a quadratic can be solved by substitution.
Let m = x2Use a ‘let’ statement to define a variable for substitution.
9m2 - 25m + 16 = 0Substitute into the original equation.
m = 4, or m = 9Solve using any method.
x2 = 4, or x2 = 9Substitute x2back into the solution.
x = ±2, or x = ± 3Solve.
For Example: Solve
Solution: A variation on the last one, remember x =.
Let m =Use a ‘let’ statement to define the variable.
m2 + 3m - 18 = 0Substitute into the original equation.
m = -6, or m = 3Solve using any method.
= -6, or = 3Substitute back into the solution.
x = 36 or x = 9Solve.
When you check this one, you will notice that it only works if you use = -6 and =+3.
Quadratics may be disguised using the methods listed above, as well as variations and combinations of them.
Practice Questions 4: Equations of degree other than 2.
1.Solve the following;
a. 15x3 -3x2 - 12x = 0b. x + 5 - 14 = 0
c. x5 - 5x3 - 36x = 0d. x4 - 34x2 + 225 = 0
e. (x - 3)2 + 7(x - 3) + 10 = 0f. 6x-2 + 11x-1 - 10 = 0
g. 6x - 5 - 1 = 0h. 15(2x - 3)2 - 7(2x - 3) - 2 = 0
i. 4x-2 + 25x-1 + 25 = 0
Solving Quadratic Inequalities
The solution of an inequality is usually a range of numbers. This can be determined by solving the inequality;
For Example: Solve 3x + 1 ≤2
Solution: Solve the inequality as you would solve the same equation.
3x - 1 ≤2
3x ≤3Add 1 to each side.
x ≤1Divide both sides by 3.
Graph on a number line.
In an inequality of degree 2 the range of the solution is determined by both of the possible solutions.
For Example: Solve x2 - 4 > 0
Solution: Solve the inequality as you would solve the same equation.
x2 - 4 > 0
x2 > 4Add 4 to both sides.
|x| >Roots have 2 solutions. They determine the
boundaries of the solution.
x = ±2The boundaries.
Graph showing the boundaries.
02 - 4 = -4 < 0Using 0 as a test point shows that the middle
does not work.
(-3)2 - 4 = 1 > 0Using -3 as a test point shows that the lower
range is part of the solution.
(3)2 - 4 = 1 > 0Using 3 as a test point shows that the upper
range is part of the solution.
Graph of the solution.
-2 > x > 2The solution as an inequality.
Notice that the two solutions separate the number line into three areas. In the example above the areas were; x < -2, -2 < x < 2, and x > 2.
Once the solutions to the quadratic are known, we can choose test points in each section of the number line. These test points can be substituted into the original equation to determine the solution set.
For Example: Solve; x2 - 2x - 8 < 0
Solution; Solve the inequality as you would solve the same equation. This can be done by factoring, or the quadratic formula.
(x - 4)(x + 2) < 0The Factored equation.
x = -2, x = 4 are the key points or boundaries of the solution.
Number line showing the boundaries.
Area of Line / x < -2 / -2 < x < 4 / x > 4Point / x = -3 / x = 0 / x = 5
Calculation / x = (-3 - 4)(-3 + 2) / x = (0 - 4)(0 + 2) / x = (5 + 4)(5 + 2)
Result / X = 7 / x = -8 / x = 7
Truth Table / x0 False / x < 0 True / x < 0 False
The Solution: -2 < x < 4
Notice that the question of greatest interest is;
“Is the value of x at the test point positive (x > 0) or negative (x < 0)?”
We can, therefore, simplify our chart as follows;
If we use the quadratic formula to solve the equation, we can use this chart by substituting the solutions into the factored form.
For Example: Solve; x2 - 3x - 5 ≤ 0.
Solution: Solve the quadratic using the quadratic formula.
The solution from the quadratic formula.
The ‘factored’ equation. Notice the ‘-’ signs
(x + 1.19)(x - 4.19)In decimal form. (It makes the chart easier.)
x < -1.19 / -1.19 < x < 4.19 / x > 4.19(x + 1.19) / - / - / +
(x - 4.19) / - / + / +
(x + 1.19)(x - 4.19) / + / - / +
x 0 ? / False / True / False
The solution: