Solve the following equations for the unknown x:
Problem 1 :
5 = 7x - 16
Solution:
5 = 7x - 16
7x = 5 + 16
7x = 21
x = 21/7
x = 3
Problem 2 :
2x - 3 = 5 - x
Solution:
2x - 3 = 5 - x
2x + x = 5 + 3
3x = 8
x = 8/3
Multiply the indicated polynomials and simplify.
Problem 3 :
(x + 1)(x^{2} - x + 1)
Solution:
= (x + 1)(x^{2} - x + 1)
= x^{3} - x^{2} + x + x^{2} - x + 1
= x^{3} + 1
Problem 4 :
(x - 2)(x + 2)
Solution:
= (x - 2)(x + 2)
= x^{2} + 2x - 2x - 4
= x^{2} - 4
Problem 5 :
(x - 2)(x - 2)
Solution:
= (x - 2)(x - 2)
= x^{2} - 2x - 2x + 4
= x^{2} - 4x + 4
Find the domain of each of the following functions.
Problem 6 :
Solution:
√(1 + x) = 0
x = -1
x ≠ -1
1 + x ≥ 0
x ≥ -1
x ≠ -1 and x ≥ -1
x > -1
Domain = (-1, ∞)
Problem 7 :
Solution:
1 + x^{2} ≠ 0
x ∈ R
Domain = (-∞, ∞)
Problem 8 :
Given that f(x) = x^{2} - 3x + 4, find and simplify f(3), f(a), f(-t) and f(x^{2} + 1).
Solution:
f(x) = x^{2} - 3x + 4
f(3) = 3^{2} - 3(3) + 4
= 9 - 9 + 4
f(3) = 4
f(a) = a^{2} - 3(a) + 4
= a^{2} - 3a + 4
By using quadratic formula,
f(-t) = (-t)^{2} - 3(-t) + 4
= t^{2} + 3t + 4
By using quadratic formula,
f(x^{2} + 1) = (x^{2} + 1)^{2} - 3(x^{2} + 1) + 4
= x^{4} + 2x^{2} + 1 - 3x^{2} - 3 + 4
= x^{4} - x^{2} + 2
Factor the following quadratics.
Problem 9 :
x^{2} - x - 20
Solution:
x^{2} - x - 20 = 0
x^{2} + 4x - 5x - 20 = 0
x(x + 4) - 5(x + 4) = 0
(x - 5)(x + 4) = 0
x - 5 = 0 or x + 4 = 0
x = 5 or x = -4
Problem 10 :
-2x^{2} + 7x + 15
Solution:
-2x^{2} + 7x + 15 = 0
-(2x^{2} - 7x - 15) = 0
-(2x^{2} + 3x - 10x - 15) = 0
-(x(2x + 3) - 5(2x + 3)) = 0
-(x - 5)(2x + 3) = 0
x - 5 = 0 or 2x + 3 = 0
x = 5 or x = -3/2
Problem 11 :
x2 - 2
Solution:
x^{2} - 2 = 0
x^{2} = 2
x = ±√2
Solve the following quadratic equations in three ways.
1) factor
2) quadratic
3) completing the square.
Problem 12 :
-x^{2} - 3x - 2 = 0
Solution:
Factor:
-x^{2} - 3x - 2 = 0
-(x^{2} + 3x + 2) = 0
-(x^{2} + 2x + x + 2) = 0
-(x(x + 2) + 1(x + 2)) = 0
-(x + 1) (x + 2) = 0
x + 1 = 0 or x + 2 = 0
x = -1 or x = -2
Quadratic formula :
-x^{2} - 3x - 2 = 0
Completing the square method:
-x^{2} - 3x - 2 = 0
Divide by -1.
x^{2} + 3x + 2 = 0
x^{2} + 3x = -2
x^{2} + 2(x)(3/2) = -2
x^{2} + 2(x)(3/2) + (3/2)^{2} = -2 + (3/2)^{2}
(x + 3/2)^{2} = -2 + 9/4
(x + 3/2)^{2} = 1/4
Taking square root on both sides,
√(x + 3/2)^{2} = √(1/4)
x + 3/2 = ±1/2
x = 1/2 - 3/2 or x = -1/2 - 3/2
x = (1 - 3)/2 or x = (-1 - 3)/2
x = -2/2 or x = -4/2
x = -1 or x = -2
Problem 13 :
2x^{2} + 2x - 4 = 0
Solution:
2x^{2} + 2x - 4 = 0
Divide by 2,
x^{2} + x - 2 = 0
x^{2} + 2x - x - 2 = 0
x(x + 2) - 1(x + 2) = 0
(x - 1)(x + 2) = 0
x = 1 or x = -2
Quadratic formula :
2x^{2} + 2x - 4 = 0
x^{2} + x - 2 = 0
Completing the square method:
2x^{2} + 2x - 4 = 0
x^{2} + x - 2 = 0
x^{2} + x = 2
x2 + 2(x)(1/2) = 2
x^{2} + 2(x)(1/2) + (1/2)^{2} = 2 + (1/2)^{2}
(x + 1/2)^{2} = 2 + 1/4
(x + 1/2)^{2} = 9/4
Take square root on both sides.
√(x + 1/2)^{2} = √9/4
(x + 1/2) = ±3/2
x + 1/2 = 3/2 or x + 1/2 = -3/2
x = 3/2 - 1/2 or x = -3/2 - 1/2
x = (3 - 1)/2 or x = (-3 - 1)/2
x = 2/2 or x = -4/2
x = 1 or x = -2
Solve the following smorgasbord of equations and inequalities.
Problem 14 :
√x = √(2x - 1)
Solution:
√x = √(2x - 1)
Take square on both sides,
x = 2x - 1
x - 2x = -1
x = 1
Checking the original equation:
LHS = √1
RHS = √(2(1) - 1) = √1
Problem 15 :
√x^{2} - 3 = √2x
Solution:
√x^{2} - 3 = √2x
Take square root on both sides,
x^{2} - 3 = 2x
x^{2} - 2x - 3 = 0
(x - 3)(x + 1) = 0
x = 3 or x = -1
Checking the original equation:
LHS = √(3)^{2} - 3
= √(9 - 3)
= √6
= 2√3 = RHS
Problem 16 :
|x - 5| = 4
Solution:
|x - 5| = 4
x - 5 = 4 or x - 5 = -4
x = 9 or x = 1
Checking the original equation:
|9 - 5| = | 4| = 4
|1 - 5| = |-4| = 4
Problem 17 :
2x + 4 ≥ 3
Solution:
2x + 4 ≥ 3
2x ≥ -1
x ≥ -1/2
If x ≥ -1/2 that will imply that 2x + 4 ≥ 3.
So, the interval [-1/2, ∞).
Problem 18 :
-2x + 4 ≥ 3
Solution:
-2x + 4 ≥ 3
-2x ≥ -1
x ≤ 1/2
If x ≤ 1/2 this will imply that -2x + 4 ≥ 3.
So, the interval (-∞, 1/2].
Problem 19 :
Solution:
Checking the original equation:
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