SAT QUADRATICS PRACTICE PROBLEMS

Problem 1 :

In the xy plane, what is the distance between the two x-intercepts of the parabola 

y = x² - 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

x² - 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 x² + 4x + 2 = 0 ?

(a)  -2 ± √2      (b)  2 ± √2   (c) -2 ± 2√2    (d)  -4 ± 2√2

Solution :

Problem 3 :

If a < 1 and 2a² - 7a + 3 = 0, what is the value of a ?

Solution :

2a² - 7a + 3 = 0

Using factoring method :

2a² - 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 :

3x² + 10x = 8

If a and b are two solutions to the equation above and a > b, what is the value of b² ?

Solution :

3x² + 10x = 8

3x² + 10x - 8 = 0

3x² + 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 

b² = (-4)²

b² = 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) + 3² = 4x + 5

4x² - 12x + 9 - 4x - 5 = 0

4x² - 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 = x² + 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 = x² + cx  -----(2)

(1) = (2)

-3 = x² + cx 

x² + cx + 3 = 0

Applying c as -4 from (1),

x² - 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 - t², result in the maximum value of v ?

Solution :

v = 5t - t²

Writing the given quadratic function in vertex form, we get

v = -(t² - 5t)

v = -(t² - 5t)

Maximum value is at 2.5.

Problem 10 :

P = m² - 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 = m² - 100 m - 120000

When p = 0 (there is no lose)

m² - 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 = ax² + 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 = ax² + 4x - 4 -----(2)

To find check if the system of equations is having one real solution.

(1) = (2)

-3 = ax² + 4x - 4

ax² + 4x - 4 + 3 = 0

ax² + 4x - 1 = 0

When a = -4, -4x² + 4x - 1 = 0

b² - 4ac = 0

When the roots are real and equal

4²-4(a)(-1) = 0

16 + 4a = 0

4a = -16

a = -4

Problem 12 :

f(x) = -x² + 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) = -x² + 6x + 20

f(x) = -(x² - 6x - 20)

f(x) = -(x² - 2(x) (3) + 3² - 3² - 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 + x₂)/2 = 5

3 + x₂ = 10

x₂ = 10 - 3

x₂ = 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 = x² + bx + 5

at the point (3, k). If b is a constant, what is the value of k ?

Solution :

y = 2x + b ----(1)

y = x² + 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 = x² + bx + 5

Here x = 3

6 + b = 3² + 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.