Problem 1 :
(3x + 2y)^{2}
In the expression above can be written as ax^{2} + bxy + cy^{2}, where a, b and c are constants, where a + b + c ?
Solution :
(3x + 2y)^{2}
Using the algebraic identity,
(a + b)^{2} = a^{2} + 2ab + b^{2}
Here a = 3x and b = 2y
(3x + 2y)^{2} = (3x)^{2} + 2(3x)(2y) + (2y)^{2}
= 9x^{2} + 12xy + 4y^{2}
By comparing with ax^{2} + bxy + cy^{2}, we get
a = 9, b = 12 and c = 4
a + b + c = 9 + 12 + 4
= 25
So, the answer is 25.
Problem 2 :
x^{2} + kx + 9 = (x + a)^{2}
In the equation above, k and a are positive constants. If the equation is true for all values of x, what is the value of k ?
a) 0 b) 3 c) 6 d) 9
Solution :
x^{2} + kx + 9 = (x + a)^{2}
Applying the algebraic identity, (a + b)^{2 }= a^{2} + 2ab + b^{2}
x^{2} + kx + 9 = x^{2} + 2ax + a^{2}
Comparing the coefficients, we get
a^{2 }= 9 and k = 2a
a = 3
Applying the value of a, we get
k = 2(3)
k = 6
So, the value of k is 6.
Problem 3 :
If (x + 3) (x - 3) = 91, what is the value of x^{2} ?
Solution :
(x + 3) (x - 3) = 91
Using the algebraic identity,
(a + b)(a - b) = a^{2} - b^{2}
x^{2} - 3^{2} = 91
x^{2} - 9 = 91
x^{2} = 91 + 9
x^{2} = 100
So, the value of x^{2} is 100.
Problem 4 :
If (c + d) = -5 and (c - d) = -12, then what is the value of c^{2} - d^{2}?
Solution :
This problem can be done in two ways,
1) using the method of elimination
2) using algebraic identity
By using algebraic identity, we can do the problem simply
(a + b)(a - b) = a^{2} - b^{2}
(c + d) (c - d) = -5 (-12)
c^{2} - d^{2 }= 60
So, the value of c^{2} - d^{2 }is 60.
Problem 5 :
If (mx + c) (nx + 3) = 12x^{2} + 5x - 3 for all values of x, where m, n and c are constants, what is the value of m + n ?
a) 7 b) 8 c) 12 d) 13
Solution :
(mx + c) (nx + 3) = 12x^{2} + 5x - 3
mnx^{2} + 3mx + cnx + 3c = 12x^{2} + 5x - 3
mnx^{2} + (3m + cn)x + 3c = 12x^{2} + 5x - 3
Comparing the coefficients of x^{2}, x and constant.
mn = 12 -----(1)
3m + cn = 5 -----(2)
3c = -3
c = -1
Applying the value of c in (2), we get
3m + (-1)n = 5
3m - n = 5
From (1),
m = 12/n
3(12/n) - n = 5
(36/n) - n = 5
(36 - n^{2})/n = 5
36 - n^{2} = 5n
n^{2} + 5n - 36 = 0
(n + 9)(n - 4) = 0
Equating each factor to 0, we get
n + 9 = 0 and n - 4 = 0
n = -9 and n = 4
When n = -9, m = 12/(-9) ==> -4/3
When n = 4, m = 12/4 ==> 3
m + n = (-4/3) + (-9) ==> (-4-27)/3 ==> -31/3
m + n = 4 + 3 ==> 7
So, the value of m + n is 7, option a is correct.
Problem 6 :
In the equation above, a, b and c are constants. If the equation is true for all values of x, what is the value of a + b + c?
Solution :
Comparing the corresponding terms, we get
a = 30, b = 5 and c = 20
a + b + c = 30 + 5 + 20
= 55
So, the value of a + b + c is 55.
Problem 7 :
If (x + y)^{2} - (x - y)^{2} = 60 and x and y are positive integers, which of the following could be the value of x + y ?
a) 6 b) 8 c) 10 d) 12
Solution :
(x + y)^{2} - (x - y)^{2} = 60
x^{2} + 2xy + y^{2} - (x^{2} - 2xy + y^{2}) = 60
x^{2} + 2xy + y^{2} - x^{2} + 2xy - y^{2} = 60
4xy = 60
When x = 3 and y = 5, then 4xy = 4(3)(5) ==> 60
Then the values of x and y are 3 and y respectively.
x + y = 3 + 5 ==> 8
So, the value of x + y is 8.
Problem 8 :
(a + b)^{2} - (a - b)^{2}
The expression above equivalent to which of following ?
a) 2ab b) 4ab c) 4ab + 2b^{2 } d) 2a^{2} + 2b^{2}
Solution :
(a + b)^{2} - (a - b)^{2}
Using the algebraic identities,
= (a + b)^{2} - (a - b)^{2}
= (a^{2} + 2ab + b^{2}) - (a^{2} - 2ab + b^{2})
= a^{2} + 2ab + b^{2} - a^{2} + 2ab - b^{2}
= 2ab + 2ab
= 4ab
So, the answer is option b.
Problem 9 :
(x - c)^{2} = x + 3
If c = 3, what is the solution set of the equation above ?
a) {1} b) {6} c) {1, 6} d) {-3, 1, 6}
Solution :
(x - c)^{2} = x + 3
When c = 3, we get
(x - 3)^{2} = x + 3
x^{2} - 2(x) (3) + 3^{2} = x + 3
x^{2} - 6x + 9 = x + 3
x^{2} - 6x - x + 9 - 3 = 0
x^{2} - 7x + 6 = 0
Solving this quadratic equation by factoring,
(x - 6)(x - 1) = 0
Equation each factor to 0, we get
x - 6 = 0 and x - 1 = 0
x = 6 and x = 1
So, the values of x are {1, 6}, option c is correct.
Problem 10 :
x^{2} - y^{2} = 48
x + y = 12
If (x, y) is the solution to the system of equations above, what is the value of xy?
a) 28 b) 32 c) 45 d) 64
Solution :
x^{2} - y^{2} = 48 ----(1)
x + y = 12 ----(2)
Using algebraic identity,
x^{2} - y^{2} = 48
(x + y) (x - y) = 48
Applying the value of (2), we get
12 (x - y) = 48
x - y = 48/12
x - y = 4 ----(3)
(2) + (3)
x + y + x - y = 12 + 4
2y = 16
y = 8
Applying the value of y in (2), we get
x + 8 = 12
x = 12 - 8
x = 4
x y = 4(8) ==> 32
So, the value of xy is 32.
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May 21, 24 08:51 AM
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