Solve the following inequations:-
1.log x^2 to the base(2*x+3)<1
2.|x^2+3*x|+x^2-2>=0
3.(2*x+5)^1/2+(x-1)^1/2>8
log( x^2 )[(2*x+3)] < 1
Let log( x^2 )[(2*x+3)] = p ----- (1)
i.e. p < 1
Then, by definition of logs, using (1),
(2x + 3)^p = x^2
And,
(2x + 3)^p < (2x + 3)^1 (using p < 1, assuming (2x+3) > 1)
x^2 < (2x + 3)^1
x^2 – 2x – 3 < 0
(x – 3)(x + 1) < 0
x < 3 and x > -1
Answer: [-1 < x < 0], [0 < x < 3]
|x^2+3*x| + x^2 - 2 >= 0
|x^2+3*x| >= -x^2 + 2
i.e.
(x^2+3*x) >= -x^2 + 2 -(x^2+3*x) >= -x^2 + 2
2x^2 + 3x – 2 >= 0 -3x – 2 >= 0
(2x – 1)(x + 2) >= 0 3x <= -2
x >= ½ , x <= -2 x <= -2/3
x <= -2 and x <= -2/3 is satisfied by x <= -2/3
Answer: [-∞, -2/3], [1/2, ∞]
(2*x + 5)^1/2 + (x - 1)^1/2 > 8
Squaring both sides,
(2*x + 5) + (x - 1) + 2*{(2*x + 5)(x - 1)}^1/2 > 64
3x + 4 + 2*{(2*x + 5)(x - 1)}^1/2 > 64
2*{(2*x + 5)(x - 1)}^1/2 > 64 – 3x – 4
2*{(2*x + 5)(x - 1)}^1/2 > 60 – 3x
Squaring both sides,
4{2x^2 + 3x - 5} > 3600 – 360x + 9x^2
8x^2 + 12x - 20 > 3600 – 360x + 9x^2
x^2 – 372x + 3620 < 0
(x – 10)(x – 362) < 0
[10 < x < 362]
The above range is a solution to the final inequality, but the initial inequality is satisfied with [x > 10] since square root function only considers positive values.
Answer: [x > 10]