8. Solving Quadratic Equations With the Quadratic Formula
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Chapter 8
8.
Solving Quadratic Equations With the Quadratic Formula
| Student Learning Objectives: |
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Why
Solving quadratic equations becomes much more flexible when you understand the Quadratic Formula. This lesson introduces the formula, shows where it comes from, and explains how it can be used to solve quadratic equations even when factoring or graphing is not convenient. It also introduces the discriminant, which helps determine whether a quadratic equation has two, one, or no real solutions without actually solving it. By learning these tools, students can approach quadratic equations more confidently and connect algebraic solutions to real-world situations involving height, profit, area, and motion.
Lesson Settings & Tools
| | 10 Theory slides |
| | 10 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Solving Quadratic Equations With the Quadratic Formula
Slide of 10
There are several methods for solving quadratic equations. In this lesson, the Quadratic Formula will be presented, proven, and used. Additionally, a method for determining the number of solutions without actually solving the equation will be presented.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Try a few practice exercises as a warm-up!
a Simplify the numeric expression sqrt(4^2 - 4(1)(2)) by using the properties of square roots to remove the perfect-square factor.
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b Consider the quadratic function y = (x+2)^2-1. Of the following expressions, which represents the same function written in standard form?
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c Identify the coefficient b of the quadratic equation x-5=2(x-3)^2 when written in standard form ax^2+bx+c = 0.
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Challenge
Is There a Solution?
Magdalena and Diego, both huge fans of statistics, went camping to bond under the stars and talk stats. However, they realize that bears are in the area. They need to hang their food basket from a branch 15 feet above the ground. Diego figures he can throw a stone with a rope attached to it over the branch. As Diego winds up, Magdalena sheepishly snickers, "No way that works."
In wondering if Diego's throw will be a success, consider the following quadratic function that models the height, in air, of the stone's location after t seconds of being thrown. h(t) = 5(-3t^2 + 5t + 1)
Magdalena also wonders what quadratic equation represents this scenario. Help her find it. Then, without solving the equation, determine whether it is even possible to know if the stone will reach the branch.
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Discussion
The Quadratic Formula
Besides graphing, using square roots, factoring, and completing the square, there is another method for solving a quadratic equation. This method consists of using the Quadratic Formula. Check out how to derive the formula by completing the square!
Rule
Quadratic Formula
The Quadratic Formula can be used to solve a quadratic equation written in standard form ax^2+bx+c = 0.
x=- b±sqrt(b^2-4ac)/2a
Proof
The Quadratic Formula can be derived by completing the square given the standard form of the quadratic equation ax^2 + bx + c = 0. This method will be used to isolate the x-variable. To complete the square, there are five steps to follow.
1
Factor Out the Coefficient of x^2
expand_more It is easier to complete the square when the quadratic expression is written in the form x^2 + bx + c. Therefore, the coefficient a should be factored out.
ax^2+bx+c = 0 ⇕ a(x^2+b/ax+c/a) = 0
Since the equation is quadratic, the coefficient a is not equal to 0. Therefore, both sides of the equation can be divided by a.
a(x^2+b/ax+c/a) = 0 ⇕ x^2+b/ax+c/a = 0
2
Identify the Constant Needed to Complete the Square
expand_more The next step is to rewrite the equation by moving the existing constant to the right-hand side. To do so, ca will be subtracted from both sides of the equation. x^2+b/ax+ c/a = 0 ⇕ x^2+ b/ax = - c/a
The constant needed to complete the square can now be identified by focusing on the x-term, while ignoring the rest. One way to find this constant is by squaring half the coefficient of the x-term, which in this case is ba.
Note that leaving the constant as a power makes the next steps easier to perform.
3
Complete the Square
expand_more The square can now be completed by adding the constant found in Step 2 to both sides of the equation.
\begin{gathered}
x^2+\dfrac{b}{a}x = \text{-} \dfrac{c}{a} \\ \Updownarrow \\ \underbrace{x^2+\dfrac{b}{a}x + {\color{#FD9000}{\left(\dfrac{b}{2a} \right)^2}}}_\text{Perfect Square Trinomial} = \underbrace{\text{-} \dfrac{c}{a} + {\color{#FD9000}{\left(\dfrac{b}{2a} \right)^2}}}_\text{Constant}
\end{gathered}
The first three terms form a perfect square trinomial, which can be factored as the square of a binomial. The other two terms do not contain the variable x. Therefore, their value is constant.
4
Factor the Perfect Square Trinomial
expand_more The perfect square trinomial can now be factored and rewritten as the square of a binomial.
x^2+b/ax + (b/2a )^2 = - c/a + (b/2a )^2
DenomMultFracToNumber
2 * a/2= a
x^2+2(b/2a)x + (b/2a )^2 = - c/a + (b/2a )^2
CommutativePropMult
Commutative Property of Multiplication
x^2+2x(b/2a) + (b/2a )^2 = - c/a + (b/2a )^2
FacPosPerfectSquare
a^2+2ab+b^2=(a+b)^2
(x + b/2a)^2 = - c/a + (b/2a )^2
The process of completing the square is now finished.
5
Simplify the Equation
expand_more Finally, the right-hand side of the equation can be simplified.
(x + b/2a)^2 = - c/a + (b/2a )^2
▼
Simplify right-hand side
CommutativePropAdd
Commutative Property of Addition
(x + b/2a)^2 = (b/2a )^2 - c/a
PowQuot
(a/b)^m=a^m/b^m
(x + b/2a)^2 = b^2/(2a)^2 - c/a
PowProdII
(a b)^m=a^m b^m
(x + b/2a)^2 = b^2/4a^2 - c/a
ExpandFrac
a/b=a * 4a/b * 4a
(x + b/2a)^2 = b^2/4a^2 - c* 4a/a * 4a
CommutativePropMult
Commutative Property of Multiplication
(x + b/2a)^2 = b^2/4a^2 - 4ac/4a* a
ProdToPowTwoFac
a* a=a^2
(x + b/2a)^2 = b^2/4a^2 - 4ac/4a^2
SubFrac
Subtract fractions
(x + b/2a)^2 = b^2-4ac/4a^2
Now, there is only one x-term. To isolate x, it is necessary to take square roots on both sides of the equation. This results in both a positive and a negative term on the right-hand side. sqrt((x + b/2a)^2) = sqrt(b^2-4ac/4a^2) ⇕ x + b/2a = ± sqrt(b^2-4ac/4a^2) Now, the equation can be further simplified to isolate x.
x + b/2a = ± sqrt(b^2-4ac/4a^2)
▼
Solve for x
SqrtQuot
sqrt(a/b)=sqrt(a)/sqrt(b)
x + b/2a = ± sqrt(b^2-4ac)/sqrt(4a^2)
SqrtProd
sqrt(a* b)=sqrt(a)*sqrt(b)
x + b/2a = ± sqrt(b^2-4ac)/2sqrt(a^2)
SqrtPowToNumber
sqrt(a^2)=a
x + b/2a = ± sqrt(b^2-4ac)/2a
SubEqn
LHS-b/2a=RHS-b/2a
x = - b/2a ± sqrt(b^2-4ac)/2a
MoveNegFracToNum
Put minus sign in numerator
x = - b/2a ± sqrt(b^2-4ac)/2a
AddSubFrac
Add and subtract fractions
x = - b ± sqrt(b^2-4ac)/2a
Finally, the Quadratic Formula has been obtained.
x = - b ± sqrt(b^2-4ac)/2a
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Example
Solving a Quadratic Equation Using the Quadratic Formula
Magdalena will sell lottery tickets as a fundraiser to support paralympic athletes. The total profit p(x) depends on the price x of a ticket and can be modeled by using the following quadratic equation. p(x) = -2x^2 + 32x + 104 Magdalena wants to raise at least $200. However, she has not yet set the price of each lottery ticket. Help Magdalena find the smallest amount that can be charged per ticket and still make a profit of at least $200. Round the price to the nearest whole dollar (the dollar sign is not necessary).
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Hint
Since the profit should be at least $200, let p(x) be equal to 200. Then, rewrite the quadratic equation in standard form. The equation can be solved using the Quadratic Formula.
Solution
It is given that the profit for the fundraiser should be at least $200. Consider the given quadratic equation, which models the profit, and substitute 200 for p(x). Then, rewrite the obtained equation in standard form.
p(x) = -2x^2 + 32x + 104
Substitute
p(x)= 200
200 = -2x^2 + 32x + 104
SubEqn
LHS-200=RHS-200
0 = -2x^2 + 32x - 96
RearrangeEqn
Rearrange equation
-2x^2 + 32x - 96 = 0
Now, all of the coefficients in the standard form ax^2+bx+c = 0 can be determined. -2x^2 + 32x - 96 = 0 ⇕ -2x^2 + 32x + ( - 96) = 0 Therefore, a= -2, b= 32, and c= -96. The obtained equation will be solved using the Quadratic Formula. x = - b ± sqrt(b^2-4ac)/2a The values of a, b, and c will now be substituted into the formula. Find x by evaluating the right-hand side of the formula.
x = - b ± sqrt(b^2-4ac)/2a
SubstituteValues
Substitute values
x = - 32 ± sqrt(32^2-4( -2)( -96))/2( -2)
▼
Evaluate right-hand side
CalcPow
Calculate power
x=-32±sqrt(1024-4(-2)(-96))/2(-2)
MultPosNeg
a(- b)=- a * b
x=-32±sqrt(1024-(-8)(-96))/-4
MultNegNegTwoPar
(- a)(- b)=a* b
x=-32±sqrt(1024-768)/-4
SubTerm
Subtract term
x=-32±sqrt(256)/-4
CalcRoot
Calculate root
x = -32 ± 16/-4
Using the Quadratic Formula, it was obtained that the solutions for the equation are x = -32 ± 16-4. Finally, both solutions can be evaluated using a table.
| x = -32 ± 16/-4 | |
|---|---|
| x = -32 + 16/-4 | x = -32 - 16/-4 |
| x=-16/-4 | x=-48/-4 |
| x=4 | x=12 |
Since Magdalena wants the tickets to be as cheap as possible while making a profit of at least $200, the price each ticket should be $4.
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Example
Solving a Quadratic Equation Not in Standard Form
A fire nozzle attached to a hose is a device used by firefighters to extinguish fires. Consider a firefighter who is aiming water to extinguish a fire on the third floor of a building. The base of the fire is situated 22 feet above the ground.
The stream of water delivered from the fire nozzle can be modeled by the following quadratic function. h(x) = -0.008x(x-100) + 4 In this equation, x is the horizontal distance from the firefighter and h(x) is the height of the water stream. Both x and h(x) are measured in feet. Knowing that the water stream's peak is 2 feet above the base of the fire, what is the horizontal distance from the firefighter to the peak of the water stream?
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Hint
What is the height of the water stream's peak? Write a quadratic equation and solve it using the Quadratic Formula.
Solution
Consider the given situation on a coordinate plane and assume that the firefighter is standing on the y-axis. 
Since the water stream's peak is 2 feet above the fire's base, whose height is 22 feet, its height h(x) is 2+22= 24 feet. This height will now be substituted into the equation of the given quadratic function to calculate the desired distance. h(x) = -0.008x(x-100) + 4 ↓ 24 = -0.008x(x-100) + 4 The obtained quadratic equation can be solved using the Quadratic Formula. To do so, the equation must first be rewritten in standard form.
24 = -0.008x(x-100) + 4
▼
Rewrite
Distr
Distribute -0.008x
24 = -0.008x^2+0.8x + 4
SubEqn
LHS-24=RHS-24
0 = -0.008x^2+0.8x -20
RearrangeEqn
Rearrange equation
-0.008x^2 + 0.8x - 20 = 0
Next, the coefficients a, b, and c can be identified. -0.008x^2 + 0.8x - 20 = 0 ⇕ -0.008x^2 + 0.8x + (- 20) = 0 Finally, these values will be substituted into the Quadratic Formula to solve the equation for x.
x=- b±sqrt(b^2-4ac)/2a
SubstituteValues
Substitute values
x=- 0.8±sqrt(0.8^2-4( -0.008)(- 20))/2( -0.008)
▼
Evaluate right-hand side
CalcPow
Calculate power
x=-0.8±sqrt(0.64-4(-0.008)(-20))/2(-0.008)
MultPosNeg
a(- b)=- a * b
x=-0.8±sqrt(0.64-(-0.032)(-20))/-0.016
MultNegNegTwoPar
(- a)(- b)=a* b
x=-0.8±sqrt(0.64-0.64)/-0.016
SubTerm
Subtract term
x=-0.8±sqrt(0)/-0.016
CalcRoot
Calculate root
x = -0.8 ± 0/-0.016
AddSubTerms
Add and subtract terms
x = -0.8/-0.016
DivNegNeg
- a/- b=a/b
x = 0.8/0.016
CalcQuot
Calculate quotient
x = 50
It has been found that this equation has exactly one solution, x = 50. Therefore, it can be said that the firefighter is standing at a horizontal distance of 50 feet from the water stream's peak.
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Pop Quiz
Solving a Quadratic Equation Using the Quadratic Formula
Solve the quadratic equations by using the Quadratic Formula. If necessary, round the answer to two decimal places.
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Discussion
Discriminant of a Quadratic Equation
In general, quadratic equations have two, one or no real solutions. Before solving a quadratic equation, the number of real solutions can be determined by using the discriminant.
Concept
Discriminant
In the Quadratic Formula, the expression b^2 - 4ac, which is under the radical symbol, is called the discriminant.
x = - b ± sqrt(b^2-4ac)/2a
A quadratic equation can have two, one, or no real solutions. Since the discriminant is under the radical symbol, its value determines the number of real solutions of a quadratic equation.
| Value of the Discriminant | Number of Real Solutions |
|---|---|
| b^2-4ac > 0 | 2 |
| b^2-4ac = 0 | 1 |
| b^2-4ac < 0 | 0 |
Moreover, the discriminant determines the number of x-intercepts of the graph of the related quadratic function.
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Example
Determining Whether There is a Solution Without Solving
A farmer wants to build a fence around a vegetable garden. To make it simple, the farmer will build it in the shape of a rectangle. The farmer has enough wood to build a fence the length of 800 feet, including the gate.
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Hint
Let x denote the length of one side of the rectangle. Then, use the fact that the length of the fence represents the perimeter of the rectangle. All things considered, how can the area of the rectangle be calculated?
Solution
Let x and y be the side lengths of the rectangle that represents the vegetable garden. The length of the fence is the perimeter P of the rectangle, which is the sum of all four side lengths.
P=x+x+y+y ⇕ P = 2x + 2y
It is given that the perimeter of the rectangle — how much wood the farmer has — is 800 feet. By substituting 800 for P, the equation can be solved for one of the side lengths, such as y.
P = 2x+2y
Substitute
P= 800
800 = 2x+2y
▼
Solve for y
SubEqn
LHS-2x=RHS-2x
800 - 2x = 2y
DivEqn
.LHS /2.=.RHS /2.
800-2x/2 = y
WriteDiffFrac
Write as a difference of fractions
800/2 - 2x/2 = y
MovePartNumRight
a* b/c=a/c* b
800/2 - 2/2x = y
CalcQuot
Calculate quotient
400 - 1x = y
IdPropMult
Identity Property of Multiplication
400 - x = y
RearrangeEqn
Rearrange equation
y = 400 - x
The side lengths can now be placed in the diagram. It can be arbitrarily assumed that the length of the horizontal side is x. Keep in mind that both x and 400-x are measured in feet.
Next, the area of the rectangle will be calculated in terms of x. The area A of a rectangle is the product of the rectangle's length and width. A = x(400-x) The obtained formula for A is represented by a quadratic function. It is given that the farmer's desired area should be at least 50 000 square feet. Therefore, this number will be substituted for A in the formula. A = x(400-x) ↓ 50 000 = x(400-x) The above is a quadratic equation that is not written in standard form. Hence, the equation will be rewritten to determine the number of solutions. Determining the number of solutions will help find if a value for x exists so that the area of the rectangle is 50 000 square feet.
50 000 = x(400-x)
▼
Rewrite
Distr
Distribute x
50 000 = 400x - x^2
SubEqn
LHS-50 000=RHS-50 000
0 = 400x - x^2 - 50 000
CommutativePropAdd
Commutative Property of Addition
0 = - x^2 + 400x - 50 000
RearrangeEqn
Rearrange equation
- x^2 + 400x - 50 000 = 0
The equation is now in standard form. This means that the number of solutions can be determined using the discriminant. Next, the coefficients a, b, and c need to be identified. - x^2 + 400x - 50 000 = 0 ⇕ -1x^2 + 400x + (- 50 000) = 0 The variables a, b, and c can then be substituted into the discriminant b^2-4ac.
b^2 - 4ac
SubstituteValues
Substitute values
400^2 - 4( -1)(- 50 000)
▼
Evaluate
CalcPow
Calculate power
160 000 - 4(-1)(-50 000)
MultPosNeg
a(- b)=- a * b
160 000 - (- 4)(-50 000)
MultNegNegOnePar
- a(- b)=a* b
160 000 - 200 000
SubTerm
Subtract term
-40 000
Since the discriminant is less than 0, there are no solutions to the equation. Therefore, the farmer will not be able to build a fence so that the area of the vegetable garden is 50 000 square feet. This means that he will need to buy more wood. Good thing he did the math before starting to construct the fence.
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Pop Quiz
Determining the Number of Real Solutions of a Quadratic Equation
Without solving the quadratic equations, use the discriminant to determine the number of real solutions.
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Closure
Determining the Number of Solutions of a Quadratic Equation Without Solving
The challenge presented at the beginning of this lesson asked if the stone thrown by Diego will reach, over some point in time, a branch located 15 feet above the ground.
The height, in feet, of the stone thrown by Diego is modeled by the following quadratic function. h(t) = 5(-3t^2 + 5t + 1) Will the stone reach the branch? There is no need to solve any equation to answer the question.
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Hint
Substitute 15 for h(t) and identify the discriminant of the resulting quadratic equation.
Solution
Here, t represents the number of seconds that passed since Diego threw the stone. Since the branch is situated 15 feet above the ground, this number can be substituted for h(t) into the formula.
h(t) = 5(-3t^2 + 5t + 1) ↓ 15 = 5(-3t^2 + 5t + 1)
Although this quadratic equation can be solved using the Quadratic Formula, it is sufficient to find whether the equation has any solutions. To do so, the discriminant can be calculated. To calculate the discriminant, the equation needs to be rewritten in standard form.
15 = 5(-3t^2 + 5t + 1)
▼
Rewrite
Distr
Distribute 5
15 = -15t^2 + 25t + 5
SubEqn
LHS-15=RHS-15
0 = -15t^2 + 25t - 10
RearrangeEqn
Rearrange equation
-15t^2 + 25t - 10 = 0
Now, identify the coefficients a, b, and c in the obtained equation. -15t^2 + 25t - 10 = 0 ⇕ -15t^2 + 25t + ( -10) = 0 Finally, the values of the coefficients will be substituted into the discriminant.
b^2-4ac
SubstituteValues
Substitute values
25^2 - 4( -15)( -10)
▼
Evaluate
CalcPow
Calculate power
625 - 4(-15)(-10)
MultPosNeg
a(- b)=- a * b
625 - (- 60)(- 10)
MultNegNegOnePar
- a(- b)=a* b
625-600
SubTerm
Subtract term
25
Because the value of the discriminant is greater than 0, there are two solutions of the equation. Therefore, Diego and Magdalena know that the stone will reach the desired branch at two points in time.
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Solving Quadratic Equations With the Quadratic Formula
Exercise 2.1
The function p(x)=x^2-2x-15 has been graphed below along with a straight line.
The straight line passes through the vertex and one of the x-intercepts of the quadratic function. Write the equation of the line in slope-intercept form.
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To determine the equation of a line, we need to know at least two points that lie on the line. We are told that the line goes through the vertex and one of the zeros of the parabola p(x).
Zeros of the Parabola
To find the zeros, we will set p(x) equal to 0 and solve for x using the Quadratic Formula.
The quadratic function intercepts the x-axis at x_1=5 and x_2=-3.
Vertex of the Parabola
The x-coordinate of the vertex of the parabola is given by the equation of the axis of symmetry of the parabola. We can determine its equation by using the formula x_s = x_1+x_22, where x_1 and x_2 are the already found zeros.
Therefore, x=1 is the x-coordinate of the vertex. We can determine the y-coordinate of the vertex by substituting x=1 into the function rule and evaluating.
The vertex of the function is (1,-16).
Equation of the Line
We are ready to determine the slope of the line. We can see in the diagram that the line intersects the x-axis at the positive zero of the parabola. Therefore, let's substitute the coordinates of the x-intercept (5,0) and the vertex (1,-16) into the slope formula to find the slope of the line.
Now that we know the line's slope, we can determine its y-intercept by substituting it and one of the points into the slope-intercept form of an equation and solving for b.
Finally, we can write the complete equation. y = 4x-20
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Hint & Answer
S
Solution
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The number of solutions to a quadratic equation can be determined by examining the discriminant, which is included in the Quadratic Formula. x=- b±sqrt(b^2-4ac)/2a The value of the discriminant will help us determine the number of real solutions to a quadratic equation.
| Value of the Discriminant | Number of Real Solutions |
|---|---|
| b^2-4ac > 0 | Two |
| b^2-4ac = 0 | One |
| b^2-4ac < 0 | Zero |
Let's determine the values of a, b, and c in the standard form of the given quadratic equation, which we will then substitute into the formula for the discriminant. x^2 + kx + 9 = 0 ⇓ 1x^2 + kx + 9 = 0 Now we will substitute a= 1, b= k, and c= 9 into the formula for the discriminant and evaluate.
The discriminant can be described by the expression k^2-36. If we set it equal to zero and solve for k, we can determine the values of k for which the equation has only one real solution.
When k=-6 or k=6, the quadratic equation has one real solution.
In order for the equation to have no real solutions, the discriminant must be negative. This allows us to write the following inequality. k^2-36 < 0 Let's now solve for k.
Now we have a quadratic inequality, which can sometimes be difficult to solve algebraically. It is usually easier to solve these inequalities graphically. If we draw y=k^2 and y=36, we see that they intersect when k=±6.
Now we can see that y=k^2 is less than 36 when k>-6 and k<6. We can write this as the following compound inequality.
-6
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Hint & Answer
S
Solution
Consider the following graph of a quadratic function.
Which equation(s) could describe the function of the graph? A. & f(x)=x^2-4x+6 B. & f(x)=- x^2-4x+6 C.& f(x)=x^2-6x+6 D.& f(x)=x^2-10x-6 E.& f(x)=x^2+5x+8 F.& f(x)=- x^2+2x+4
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Let's first list everything we know about the graph of the quadratic function.
- The vertex of the function is a minimum.
- The graph intercepts the positive part of the y-axis.
- The graph has no x-intercepts.
- The x-coordinate of the vertex is positive.
We can use this information to exclude the potential functions one at the time. Notice that we have not been given any points the parabola passes through. We only need to identify which functions match our criteria.
Minimum Value
A parabola with a minimum value opens upwards has a positive leading coefficient. Since our parabola opens upwards, this means that we can exclude B and F, where the leading coefficient is negative. & A. f(x)=x^2-4x+6 & B. f(x)= - x^2-4x+6 & * & C. f(x)=x^2-6x+6 & D. f(x)=x^2-10x-6 & E. f(x)=x^2+5x+8 & F. f(x)= - x^2+2x+4 & *
Positive y-intercept
Since the curve intercepts the y-axis on its positive side, we know that the function must have a positive constant term. Therefore, we can exclude option D where the constant is negative. & A. f(x)=x^2-4x+6 & C. f(x)=x^2-6x+6 & D. f(x)=x^2-10x - 6 & * & E. f(x)=x^2+5x+8
No x-intercepts
To determine which of the remaining functions have no x-intercepts, we will have a look at the discriminant of the functions. First we set f(x) equal to zero and identify the values of a, b, and c.
| Function | a | b | c |
|---|---|---|---|
| A. 1x^2 - 4x+ 6=0 | 1 | -4 | 6 |
| C. 1x^2 - 6x+ 6=0 | 1 | -6 | 6 |
| E. 1x^2 + 5x+ 8=0 | 1 | 5 | 8 |
The function has no x-intercepts if the value of the discriminant is less than zero. Let's calculate the discriminate for each of the three remaining functions. Recall that the formula for the discriminant of a function is b^2-4ac. If the discriminant is greater or equal to zero, we know that the function does not model the given graph. A. & ( -4)^2-4( 1)( 6)=-8 C. & ( -6)^2-4( 1)( 6)=10 * E. & 5^2-4( 1)( 8)=-7 As we can see, the function in option C has two x-intercepts because the discriminant is greater than 0. Therefore, we can exclude this option as well. & A. f(x)=x^2-4x+6 && & E. f(x)=x^2+5x+8 &&
Coordinates of the Vertex
The x-coordinate of the vertex is given by the equation of the axis of symmetry. It can be calculated by using the following formula. x_s=- b/2a We have already identified the values of a and b for both A and E. If we substitute these values into the formula, we can determine their axes of symmetry. & A. x_s=- ( -4)/2( 1)=2 [1em] & E. x_s=- 5/2( 1)=-2.5 Now we can exclude option E as well, since its axis of symmetry is located on the negative side of the x-axis. & A. f(x)=x^2-4x+6 && & E. f(x)=x^2+5x+8 && * This leaves only option A. To confirm that the function in option A does indeed meet all of our conditions, we need to check if the y-coordinate of the vertex is positive. Therefore, let's substitute the axis of symmetry x=2 into the function rule and solve for f(x), the y-value of the vertex.
Similar to the function in the graph, the y-coordinate of the vertex is positive. This means that the only function that could represent the graph is given in option A. & A. f(x)=x^2-4x+6 && ✓ & B. f(x)=- x^2-4x+6 && * & C. f(x)=x^2-6x+6 && * & D. f(x)=x^2-10x-6 && * & E. f(x)=x^2+5x+8 && * & F. f(x)=- x^2+2x+4 && *
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Solve the following equation for x when x^2=t. x^4-5x^2+4=0
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Let's perform the variable substitution x^2=t. Before we can do that, we must rewrite x^4 in terms of x^2.
Now we have a corresponding quadratic equation that we can solve by using the Quadratic Formula.
We found that the solutions to the quadratic equation are t=4 and t=1. We now need to substitute these solutions into the substitution equation x^2=t and solve for x. That way, we will find the solutions to the original equation. x^2=1 ⇒ x=±1 x^2=4 ⇒ x=±2 We have a total of four solutions to the original equation: x=-1, x=1, x=-2, and x=2.
The given equation can also be solved by rewriting the x^2-term and factoring. Start by writing -5x^2 as -4x^2 - x^2. x^4-5x^2+4=0 ⇕ x^4 - 4x^2 - x^2 + 4 = 0 Next, notice that x^2 can be factored out from the first two terms. Also, factor out -1 from the next two terms. x^4 - 4x^2 - x^2 + 4 = 0 ⇕ x^2( x^2 - 4 ) - (x^2 - 4 ) = 0 Now, ( x^2 - 4 ) can be factored out. Then, use the difference of squares formula.
Finally, the Zero Product Property can be applied to write all four solutions to the original equation. lx+2 = 0 x-2 = 0 x+1 = 0 x-1 = 0 ⇔ lx = -2 x = 2 x = -1 x = 1
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Solving Quadratic Equations With the Quadratic Formula
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