Which of the following inequalities is equivalent to 1 ≥ x?
2 answers:
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Answer:
see explanation
Step-by-step explanation:
(a)
x² + 2x + 1 = 2x² - 2 ( subtract x² + 2x + 1 from both sides
0 = x² - 2x - 3 ← in standard form
0 = (x - 3)(x + 1) ← in factored form
Equate each factor to zero and solve for x
x + 1 = 0 ⇒ x = - 1
x - 3 = 0 ⇒ x = 3
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(b)
-
=
( multiply through by 15 to clear the fractions )
5(x + 2) - 2 = 3(x - 2) ← distribute parenthesis on both sides
5x + 10 - 2 = 3x - 6
5x + 8 = 3x - 6 ( subtract 3x from both sides )
2x + 8 = - 6 ( subtract 8 from both sides )
2x = - 14 ( divide both sides by 2 )
x = - 7
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(c) Assuming lg means log then using the rules of logarithms
log ⇔ nlogx
log x = log y ⇒ x = y
Given
log(2x + 3) = 2logx
log(2x + 3) = log x² , so
x² = 2x + 3 ( subtract 2x + 3 from both sides )
x² - 2x - 3 = 0
(x - 3)(x + 1) = 0
x = 3 , x = - 1
x > 0 then x = 3