Problem 1 :
In the xy plane, what is the distance between the two x-intercepts of the parabola
y = x^{2} - 3x - 10
(a) 3 (b) 5 (c) 7 (d) 10
Solution :
To find the the distance between two x-intercepts, first let us find the x-intercepts.
x-intercept :
Put y = 0
x^{2} - 3x - 10 = 0
(x - 5)(x + 2) = 0
x - 5 = 0 and x + 2 = 0
x = 5 and x = -2
Writing down as points, we get (5, 0) and (-2, 0)
Distance between two points :
Problem 2 :
What are the solutions to x^{2} + 4x + 2 = 0 ?
(a) -2 ± √2 (b) 2 ± √2 (c) -2 ± 2√2 (d) -4 ± 2√2
Solution :
Problem 3 :
If a < 1 and 2a^{2} - 7a + 3 = 0, what is the value of a ?
Solution :
2a^{2} - 7a + 3 = 0
Using factoring method :
2a^{2} - 6a - 1a + 3 = 0
2a (a - 3) -1(a - 3) = 0
(2a - 1) (a - 3) = 0
Equating each factor to 0, we get
2a = 1 and a = 3
a = 1/2 and a = 3
Since the value of a is < 1, we choose 1/2 as solution.
Problem 4 :
3x^{2} + 10x = 8
If a and b are two solutions to the equation above and a > b, what is the value of b^{2} ?
Solution :
3x^{2} + 10x = 8
3x^{2} + 10x - 8 = 0
3x^{2} + 12x - 2x - 8 = 0
3x(x + 4) - 2(x + 4) = 0
(3x - 2)(x + 4) = 0
3x - 2 = 0 and x + 4 = 0
x = 2/3 and x = -4
a = 2/3 and b = -4
b^{2} = (-4)^{2}
b^{2} = 16
Problem 5 :
What is the sum of solutions of (2x - 3)^{2} = 4x + 5 ?
Solution:
(2x - 3)^{2} = 4x + 5
(2x)^{2} - 2(2x) (3) + 3^{2} = 4x + 5
4x^{2} - 12x + 9 - 4x - 5 = 0
4x^{2} - 16x + 4 = 0
a = 4, b = -16 and c = 4
Sum of roots (α+β) = -b/a
α+β = 16/4
α+β = 4
Problem 6 :
y = -3 and y = x^{2} + cx
In the system of equations above, c is a constant. For which of the following values of c does the system of equations have exactly two real solutions ?
(a) -4 (b) 1 (c) 2 (d) 3
Solution :
y = -3 ----(1)
y = x^{2} + cx -----(2)
(1) = (2)
-3 = x^{2} + cx
x^{2} + cx + 3 = 0
Applying c as -4 from (1),
x^{2} - 4x + 3 = 0
(x - 1)(x - 3) = 0
x = 1 and x = 3
While applying c = -4, we get two real values of x. So, option a is correct.
Problem 7 :
At which of the following points does the line with equation
y = 4
intersect the parabola
y = (x + 2)^{2} - 5
in the xy - plane ?
(a) (-1, 4) and (-5, 4) (b) (1, 4) and (-5, 4)
(c) (1, 4) and (5, 4) (d) (-11, 4) and (7, 4)
Solution :
y = 4 -----(1)
y = (x + 2)^{2} - 5 -----(2)
(1) = (2)
4 = (x + 2)^{2} - 5
(x + 2)^{2} = 9
x + 2 = ± 3
x + 2 = 3 and x + 2 = -3
x = 1 and x = -5
When x = 1, y = 4
When x = -5, y = 4
So, the point of intersections are (1, 4) and (-5, 4).
Problem 8 :
Which of the following equations represents the parabola shown in the xy-plane above ?
(a) y = (x - 3)^{2} - 8 (b) y = (x + 3)^{2} + 8
(c) y = 2(x - 3)^{2} - 8 (d) y = 2(x + 3)^{2} - 8
Solution :
Quadratic function, in vertex form.
y = a(x - h)^{2} + k
Here (h, k) ==> (3, -8)
y = a(x - 3)^{2} - 8
The parabola passes through the points (1, 0) and (5, 0)
0 = a(1 - 3)^{2} - 8
8 = a(4)
a = 2
Applying the value of a, we get
y = 2(x - 3)^{2} - 8
Problem 9 :
For what value of t does the equation v = 5t - t^{2}, result in the maximum value of v ?
Solution :
v = 5t - t^{2}
Writing the given quadratic function in vertex form, we get
v = -(t^{2 }- 5t)
v = -(t^{2 }- 5t)
Maximum value is at 2.5.
Problem 10 :
P = m^{2} - 100 m - 120000
The monthly profit of a mattress company can be modeled by the equation above, where P is the profit in dollars, and m is the number of mattresses sold. What is the minimum number of mattresses the company must sell in a given month so that it does not lose money during that month ?
Solution :
P = m^{2} - 100 m - 120000
When p = 0 (there is no lose)
m^{2} - 100 m - 120000 = 0
(m - 400) (m + 300) = 0
m = 400 and m = -300
When 400 mattresses are sold, there is no loss.
Problem 11 :
y = -3
y = ax^{2} + 4x - 4
In the system of equations above, a is constant. For which of the following values of a does the system of equations have exactly one real solution ?
(a) -4 (b) 1 (c) 2 (d) 4
Solution :
y = -3 ----(1)
y = ax^{2} + 4x - 4 -----(2)
To find check if the system of equations is having one real solution.
(1) = (2)
-3 = ax^{2} + 4x - 4
ax^{2} + 4x - 4 + 3 = 0
ax^{2} + 4x - 1 = 0
When a = -4, -4x^{2} + 4x - 1 = 0
b^{2} - 4ac = 0
When the roots are real and equal
4^{2}-4(a)(-1) = 0
16 + 4a = 0
4a = -16
a = -4
Problem 12 :
f(x) = -x^{2} + 6x + 20
The function f is defined above. Which of the following equivalent forms of f(x) displays the maximum value of f as a constant or coefficient ?
(a) y = -(x - 3)^{2} + 11 (b) y = -(x + 3)^{2} + 29
(c) y = -(x + 3)^{2} + 11 (d) y = -(x + 3)^{2} + 29
Solution :
f(x) = -x^{2} + 6x + 20
f(x) = -(x^{2} - 6x - 20)
f(x) = -(x^{2} - 2(x) (3) + 3^{2} - 3^{2} - 20)
f(x) = -[(x - 3)^{2} - 9 - 20]
f(x) = -(x - 3)^{2} + 29
So, option b is correct.
Problem 13 :
y = a(x - .3)(x - k)
In the quadratic equation above, a and k are constants. If the graph of the equation in the xy plane is a parabola with vertex (5, -32), what is the value of a ?
Solution :
One of the x-intercept is 3.Since the x-coordinate of x is 5, in general the vertex will lie exactly middle of two x-intercepts.
Then, (3 + x_{2})/2 = 5
3 + x_{2} = 10
x_{2} = 10 - 3
x_{2} = 7
So, another x-intercept is 7.
y = a(x - .3)(x - 7)
Vertex is also one of the points of the curve.
-32 = a(5 - 3)(5 - 7)
-32 = a(2)(-2)
-4a = -32
a = 8
Problem 14 :
In the xy plane, the line y = 2x + b intersects the parabola
y = x^{2} + bx + 5
at the point (3, k). If b is a constant, what is the value of k ?
Solution :
y = 2x + b ----(1)
y = x^{2} + bx + 5 ----(2)
Since the point of intersection is at (3, k), it is common for both line and the parabola.
(1) = (2)
2x + b = x^{2} + bx + 5
Here x = 3
6 + b = 3^{2} + 3b + 5
6 + b = 14 + 3b
6 - 14 = 2b
2b = -8
b = -4
By applying the value of x, y and b in (1), we get
k = 2(3) - 4
k = 6 - 4
k = 2
So, the value of k is 2.
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