Chapter Closure
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Exercise 125 Page 463
a Rearrange the equation so all of the terms are on the left-hand side of the equation before using the Quadratic Formula.
b Apply the Zero Product Property.
c To complete the square, make sure all the variable terms are on one side of the equation and all constants are on the other side.
d Make sure the equation is written in standard form. Identify the related function and graph it.
a x=1 or x=- 7/3
b x=- 3 or x=6
c x=- 2+ sqrt(3) ≈ - 0.27 or x=- 2- sqrt(3) ≈ - 3.73
d x=- 4 or x=1.5
a We will use the Quadratic Formula to solve the given quadratic equation.
x=- b±sqrt(b^2-4ac)/2a
SubstituteValues
Substitute values
x=- 4±sqrt(4^2-4( 3)( - 7))/2( 3)
▼
Solve for x and Simplify
CalcPow
Calculate power
x=- 4±sqrt(16-4(3)(- 7))/2(3)
Multiply
Multiply
x=- 4±sqrt(16-12(- 7))/6
MultNegNegOnePar
- a(- b)=a* b
x=- 4±sqrt(16+84)/6
AddTerms
Add terms
x=- 4±sqrt(100)/6
CalcRoot
Calculate root
x=- 4± 10/6
FactorOut
Factor out 2
x=2(- 2± 5)/6
CancelCommonFac
Cancel out common factors
x=- 2± 5/3
| x=- 2± 5/3 | |
|---|---|
| x_1=- 2+ 5/3 | x_2=- 2- 5/3 |
| x_1=3/3 | x_2=- 7/3 |
| x_1=1 | x_2=- 7/3 |
Using the Quadratic Formula, we found that the solutions of the given equation are x_1=1 and x_2=- 73.
Factoring
Let's start by rewriting the middle term as a sum of two terms.x^2-3x-18=0
WriteSum
Write as a sum
x^2-6x+3x-18=0
x(x-6)+3(x-6)=0
FactorOut
Factor out (x-6)
(x+3)(x-6)=0
Solving
To solve this equation we will apply the Zero Product Property.(x+3)(x-6)=0
ZeroProdProp
Use the Zero Product Property
lcx+3=0 & (I) x-6=0 & (II)
SubEqn
(I): LHS-3=RHS-3
lx=- 3 x-6=0
AddEqn
(II): LHS+6=RHS+6
lx_1=- 3 x_2=6
c We want to solve the quadratic equation by completing the square. To do so, we will start by rewriting the equation so all terms with x are on one side of the equation and all constants on the other side.
d We are asked to solve the given quadratic equation. We will solve it by graphing. There are three steps to solving a quadratic equation by graphing.
- Write the equation in standard form, ax^2+bx+c=0.
- Graph the related function f(x)=ax^2+bx+c.
- Find the x-intercepts, if any.
Equation:& 2x^2+5x-12=0 Related Function:& f(x)=2x^2+5x-12
Graphing the Related Function
To draw the graph of the related function written in standard form, we must start by identifying the values of a, b, and c. f(x)=2x^2+5x-12 ⇕ f(x)=2x^2+5x+(- 12) We can see that a=2, b=5, and c=- 12. Now, we will follow three steps to graph the function.
- Find the axis of symmetry.
- Make a table of values using x values around the axis of symmetry.
- Plot and connect the points with a parabola.
Finding the Axis of Symmetry
The axis of symmetry is a vertical line with equation x=- b2a. Since we already know the values of a and b, we can substitute them into the formula. The axis of symmetry of the parabola is the vertical line with equation x=- 54.Making the Table of Values
Next, we will make a table of values using x-values around the axis of symmetry x=- 54.
| x | 2x^2+5x-12 | f(x) |
|---|---|---|
| - 3 | ( - 3)^2+5( - 3)-12 | - 9 |
| - 2 | ( - 2)^2+5( - 2)-12 | - 14 |
| - 54 | ( - 54)^2+5( - 54)-12 | - 15.125 |
| 0 | 0^2+5( 0)-12 | - 12 |
| 1 | 1^2+5( 1)-12 | - 5 |
Plotting and Connecting the Points
We can finally draw the graph of the function. Since a=2, which is positive, the parabola will open upwards. Let's connect the points with a smooth curve.
Finding the x-intercepts
Let's identify the x-intercepts of the graph of the related function.
We can see that the parabola intersects the x-axis twice. The points of intersection are ( - 4,0) and ( 1.5,0). Therefore the given equation has two solutions, x= - 4 and x= 1.5.
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