Solve log 1/2 (2x^2 + x + 1) >= -1
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log 1/2 (2x^2 + x + 1 ) >=-1
==> 2x^2 + x +1 >= (1/2)^-1
==> 2x^2 + x + 1 >= 2
Now subtract 2 from both sides:
==> 2x^2 + x -1 >= 0
Now let us factorise:
==> (2x -1)(x+1) >= 0
==> (2x-1) > 0 and (x+1) >0
==> x > 1/2 and x> -1
==> x belongs to (1/2, inf.)
Also,
(2x-1) < 0 and (x+1) <0
==> x < 1/2 and x < -1
==> x belongs to (-inf, -1)
Then the solution is:
x belongs to (-inf, -1) U (1/2, inf)
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log1/2 (2x^2+x+1) >= -1.
Solution:
The base of logarithm is (1/2).
Taking antilog , we get:
2x^2+x+1 > = (1/2)^-1 = 1/(1/2) = 2
2x^2+x+1 > = 2
2x^2 +x+1-2 >= 0
2x^2 +x-1 > 0
(2x-1)(x+1) > 0
2x -1 > 0 Or x < -1.
x >1/2 Or x < -1.
Because the logarithm has a subunit base, namely 1/2, the logarithmic function is decreasing.
We'll re-write the equation:
log 1/2 (2x^2 + x + 1) >= -1*1
log 1/2 (2x^2 + x + 1) >= -1*log 1/2 (1/2)
log 1/2 (2x^2 + x + 1) >= log 1/2 (1/2)^-1
log 1/2 (2x^2 + x + 1) >= log 1/2 (2)
2x^2 + x + 1 =< 2
We'll subtract 2 both sides:
2x^2 + x + 1 - 2 =< 0
2x^2 + x - 1 =< 0
We'll calculate the roots of the equation:
2x^2 + x - 1 = 0
We'll apply the quadratic formula:
x1 = [-1 + sqrt(1 + 8)] / 4
x1 = (-1 + 3) / 4
x1 = 2/4
x1 = 1/2
x2 = (-1 - 3) / 4
x2 = -1
The expression is negative over the interval [-1 , 1/2] and is positive for x< -1 and x> 1/2.
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