Chapter Closure
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Exercise 112 Page 206
a
To solve the equation ax^2+bx+c=0, use the Quadratic Formula.
b
Isolate the y-term on one side of the equation.
c
Start with using the Distributive Property.
d
Start with using the Distributive Property.
a
y=-3+sqrt(65)/4 y=-3-sqrt(65)/4
b
y=2/3x+5
c
y=-1/2 and y=-3
d
y=1/2
a
Let's start with gathering all terms on one side of the equation.
2y^2+3y=7 ⇒ 2y^2+3y-7=0
Now we have a quadratic equation in terms of only the y-variable.
y=- b±sqrt(b^2-4ac)/2a
SubstituteValues
Substitute values
y=- 3±sqrt(3^2-4( 2)( -7))/2( 2)
▼
Solve for y
CalcPow
Calculate power
y=-3±sqrt(9-4(2)(-7))/2(2)
MultNegPos
(- a)b = - ab
y=-3±sqrt(9-8(-7))/2(2)
MultNegNegOnePar
- a(- b)=a* b
y=-3±sqrt(9+56)/2(2)
AddTerms
Add terms
y=-3±sqrt(65)/2(2)
Multiply
Multiply
y=-3±sqrt(65)/4
This result tells us that we have two solutions for y. One of them will use the positive sign, and the other one will use the negative sign.
b
To solve an equation, we should isolate the y- term on one side of the equation using the Properties of Equality. In this case, we need to start by using the Distributive Property to simplify the left-hand side of the equation.
Now we can continue to solve using the Properties of Equality.
c
Let's start with using the Distributive Property to simplify the left-hand side of the equation.
y(2y+1)+3(2y+1)=0
Distr
Distribute y
2y^2+y+3(2y+1)=0
Distr
Distribute 3
2y^2+y+6y+3=0
AddTerms
Add terms
2y^2+7y+3=0
Now we have a quadratic equation in terms of only the y-variable.
y=- b±sqrt(b^2-4ac)/2a
SubstituteValues
Substitute values
y=- 7±sqrt(7^2-4( 2)( 3))/2( 2)
▼
Solve for y
CalcPow
Calculate power
y=-7±sqrt(49-4(2)(3))/2(2)
MultNegPos
(- a)b = - ab
y=-7±sqrt(49-8(3))/2(2)
MultNegPos
(- a)b = - ab
y=-7±sqrt(49-24)/2(2)
SubTerm
Subtract term
y=-7±sqrt(25)/2(2)
Multiply
Multiply
y=-7±sqrt(25)/4
CalcRoot
Calculate root
y=-7± 5/4
This result tells us that we have two solutions for y. One of them will use the positive sign, and the other one will use the negative sign.
| y=-7±5/4 | |
|---|---|
| y_1=-7+5/4 | y_2=-7-5/4 |
| y_1=-2/4 | y_2=-12/4 |
| y_1=-1/2 | y_2=-3 |
The solutions are y=- 12 and y=-3.
d
Let's start with using the Distributive Property to simplify the right-hand side of the equation.
-4y-1=4y(y-2)
Distr
Distribute 4y
-4y-1=4y^2-8y
AddEqn
LHS+4y+1=RHS+4y+1
0=4y^2-4y+1
RearrangeEqn
Rearrange equation
4y^2-4y+1=0
Now we have a quadratic equation in terms of only the y-variable.
y=- b±sqrt(b^2-4ac)/2a
SubstituteValues
Substitute values
y=-( -4)±sqrt(( -4)^2-4( 4)( 1))/2( 4)
y=4/8
ReduceFrac
a/b=.a /4./.b /4.
y=1/2
This result tells us that we have one solution, y= 12.
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