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Basic Algebra - Quadratic Equations

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For 1-3 solve the equation using the square root property
1. (m+1)^2 = -1
2. (p-1)^2 = 12
3. (t+5)^2 = -18
For 4 & 5, find the value of n such that the expression is a perfect square trinomial, then factor it.
4. X^2 -9X+n
5. Z^2 -2/5z +n
For 6-8, solve the equation by completing the square and applying the square root property
6. 3X^2 + 2X +1 = 0
7. W^2 + 4W = -13
8. b^2 + 7/2b = 2
For 9-11, write the function in the vertex form of a parabola and give the coordinates of the vertex, X-intercepts, y- intercepts and sketch a graph by hand.
9. P(x) = -5X^2 -10X -13
10. Z(x) = X^2 -6X +7
11. q(x) = -3X^2 -24X -54

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Solution Preview

For 1-3 solve the equation using the square root property
(m+1)2 = -1
Solution:
(m+1)^2=-1
Taking square root on both sides:
√((m+1)^2 )=√(-1)
Or,(m+1)=±√(-1)
Or,m=-1±√(-1)

(p-1)2 = 12
Solution:
(p-1)^2=12
Taking square root on both sides:
√((p-1)^2 )=√12
Or,(p-1)=±2√3
Or,p=1±2√3

(t+5)2 = -18
Solution:
(t+5)^2=-18
Taking square root on both sides:
√((t+5)^2 )=√(-18)
Or,(t+5)=±√(-18)
Or,t=-5±√(-18)

For 4 & 5, find the value of n such that the expression is a perfect square trinomial, then factor it.
X2 ...

Solution Summary

The expert solves an equation using the square root property. The value of n such that the expression is a perfect square trinomial is given.

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See Also This Related BrainMass Solution

Basic Algebra - Relations and functions

#28: y=-2x2+3

x y=-2x2+3 y
0 y=-2(0)2+3 3
1 y=-2(1)2+3 1
2 y=-2(2)2+3 -5
-1 y=-2(-1)2+3 1
-2 y=-2(-2)2+3 -5

• The graph of the relation is  (x)=-2x2+3.

• My 5 points for this equation is as follows:
(0,3)
(1,1)
(2,-5)
(-1,1)
(-2,-5)

• The key points and general shape of this graph is upward and the vertex is 3 located on the y-intercept. This is a parabola.

• The domain is (-,).

• The range is (-, 3].

• This graph is a function because each y value equals one x value. "If the value of the variable y is determined by the value of the variable x, then y is a function of x. So is a function of means is uniquely determined by" (Dugopolski, 2012, p. 690).

• The vertical line test passes through any x value and it only crosses that line once.

#36: y=2(x)+1
x y=2(x)+1 y
0 y=2(0)+1 1
1 y=2(1)+1 3
2 y=2(2)+1 3.82
4 y=2(4)+1 5
6 y=2(6)+1 5.89
• The graph of the relation is  (x)=2(x)+1.

• My 5 points for this equation is as follows:
(0,1)
(1,3)
(2,3.82)
(4,5)
(6,5.89)

• This graph is a function because each y value equals one x value.

• The general shape of this graph is a curve that plateau. There is no vertex.

• I selected problem 36 to be shifted 3 units upward and 4 units to the left.
y=2(x)+1
y=2(x+4)+1+3
y=2(x+4)+4

• This is the transformation of the function. I added 3 outside the radical and added 4 inside the radical.

Page 709 #28
This problem is an example of a quadratic function. We know this because it is in the form f(x)=ax2+bx+c where a, b, and c are real numbers and a is not equal to 0. In this case, we will graph the given function by plotting enouh points to figure out the shape of the graph.

Y = -2x2+3 original equation.
X Y In this portion, I just plugged in the values under the x column
-2 -5 into the equation above and found the corresponding solutions
-1 1 for y. These are the ordered pairs that will appear on the
0 3 graph.
1 1
2 -5
This is a function because each value for y only has one x value and passes a vertical line test. The vertex for this parabola is [0,3], the domain is {[-∞, ∞]}, and the range is {[-∞,3]}.

Page 709 #36
This problem is a square root function because it comes in the form f(x)= √x.

Y = 2√(x)+1 original equation.
X Y In this portion, I just plugged in the values under the x column
0 1 into the equation above and found the corresponding solutions
1 3 for y. These are the ordered pairs that will appear on the graph.
4 5
9 7
This is a function because each value for y only has one x value and passes a vertical line test. The domain for this graph is {[0, ∞]} and the range is [{1, ∞}].
Had this line been shifted up three and left four a transformation in the graph and equation would occur. (x)+ c represents a line shifting up c units and (x+c) represents a line shifting left c units. The change to the equation is shown below.
Y= 2√(x )+1
Y= 2√(x+4) +1+3
Y= 2√(x+4) +4

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