Problem 11 :
Which equation has exactly one real solution?
A. 4x2 - 12x - 9 = 0 B. 4x2 + 12x + 9 = 0
C. 4x2 - 6x - 9 = 0 D. 4x2 + 4x + 9 = 0
Solution:
A general quadratic equation can be represented as
ax2 + bx + c = 0
A.
D = b2 - 4ac
4x2 - 12x - 9 = 0
D = (-12)2 - 4(4)(-9)
= 144 + 144 = 288
D > 0
So, equation have two real and distinct solutions.
B.
D = b2 - 4ac
4x2 + 12x + 9 = 0
D = (12)2 - 4(4)(9)
= 144 - 144
D = 0
So, equation has exactly one real solution.
So, option (B) is correct.
Problem 12 :
The sum of two numbers is 24. The sum of the squares of the two numbers is 306. What is the product of the two numbers?
A. 119 B. 128 C. 135 D. 144
Solution:
x + y = 24 ---> (1)
x2 + y2 = 306 ---> (2)
Equation (1) squaring on both sides,
(x + y)2 = 242
x² + y² + 2xy = 576
306 + 2xy = 576
2xy = 576 - 306
2xy = 270
xy = 270/2
xy = 135
So, option (C) is correct.
Problem 13 :
The heights of two different projectiles after they are launched are modeled by f(x) and g(x). The function f(x) is defined as f(x) = -16x2 + 42x + 12. The table contains the values for the quadratic function g.
What is the approximate difference in the maximum heights achieved by the two projectiles?
A. 0.2 feet B. 3.0 feet C. 5.4 feet D. 5.6 feet
Solution :
y = ax2 + bx + c
The quadratic function passes through the points (0, 9) (1, 33) and (2, 25) respectively.
By applying the point (0, 9), we get
9 = a(0)2 + b(0) + c
c = 9
By applying the point (1, 33), we get
33 = a(1)2 + b(1) + 9
33 = a + b + 9
a + b = 33 - 9
a + b = 24 ----(1)
By applying the point (2, 25), we get
25 = a(2)2 + b(2) + 9
25 - 9 = 4a + 2b
4a + 2b = 16
2a + b = 8 ----(2)
(1) - (2)
-a = 16, a = -16
b = 24 + 16
b = 40
So, the required equation is g(x) = -16x2 + 40x + 9
Calculating the maximum value of f(x) :
f(x) = -16x2 + 42x + 12
x = -b/2a
x = 42/32
x = 1.31
f(1.31) = -16(1.31)2 + 42(1.31) + 12
f(1.31) = 39.56
Calculating the maximum value of g(x) :
g(x) = -16x2 + 40x + 9
x = -b/2a
x = 40/32
x = 1.25
g(1.25) = -16(1.25)2 + 40(1.25) + 9
g(1.25) = 34
f(1.31) - g(1.25) = 5.56 ft
So, approximately 5.6 ft. Option D.
Problem 14 :
Which expression is equivalent to -3x(x - 4) - 2x(x + 3)?
(A) -x2 - 1 (B) -x2 + 18x (C) -5x2 - 6x (D) -5x2 + 6x
Solution:
= -3x(x - 4) - 2x(x + 3)
= -3x2 + 12x - 2x2 - 6x
= -5x2 + 6x
So, option (4) is correct.
Problem 15 :
The length f a rectangle is 3 inches more than its width. The area of the rectangle is 40 square inches. What is the length, in inches, of the rectangle?
(A) 5 (B) 8 (C) 8.5 (D) 11.5
Solution:
Given A = 40 in2
l = w + 3
Area = l × w
40 = (w + 3) × w
40 = w2 + 3w
w2 + 3w - 40 = 0
(w + 8) (x - 5) = 0
w = -8 or w = 5
Since length can't be in negative.
length = w + 5
= 5 + 3
length = 8
So, option (2) is correct.
Problem 16 :
Which expressions represents 36x2 - 100y6 factored completely?
(A) 2(9x + 25y3)(9x - 25y3) (B) 4(3x + 5y3)(3x - 5y3)
(C) (6x + 10y3)(6x - 10y3) (D) (18x + 50y3)(18x - 50y3)
Solution:
= 36x2 - 100y6
= 4(9x² - 25y6)
= 4[(3x)² - (5y³)²]
Using algebraic identity,
= 4(3x + 5y3)(3x - 5y3)
So, option (2) is correct.
Problem 17 :
What are the roots of the equation x2 - 5x + 6 = 0?
(A) 1 and -6 (B) 2 and 3 (C) -1 and 6 (D) -2 and -3
Solution:
x2 - 5x + 6 = 0
x2 - 2x - 3x + 6 = 0
x(x - 2) - 3(x - 2) = 0
(x - 2)(x - 3) = 0
x - 2 = 0 and x - 3 = 0
x = 2 and x = 3
So, option (2) is correct.
Problem 18 :
Which expression is equivalent to 64 - x2 ?
(A) (8 - x)(8 - x) (B) (8 - x)(8 + x) (C) (x - 8)(x - 8)
(D) (x - 8)(x + 8)
Solution:
= 64 - x2
= 8² - x²
Using algebraic identity,
(a² - b²) = (a - b)(a + b)
= (8 - x)(8 + x)
So, option (2) is correct.
Problem 19 :
The equation of the axis of symmetry of the graph of y = 2x² - 3x + 7 is
Solution:
y = 2x² - 3x + 7
a = 2
b = -3
c = 7
So, option (A) is correct.
Problem 20 :
The roots of the equation 3x2 - 27x = 0 are
(A) 0 and 9 (B) 0 and -9 (C) 0 and 3 (D) 0 and -3
Solution:
3x2 - 27x = 0
3x(x - 9) = 0
3x = 0 and x - 9 = 0
x = 0 and x = 9
So, option (A) is correct.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM