Question

Solutions of the inequality [(4x²-3x-1)^1/2]>= 2x-3 are?

Answer 1

$[(4x^2-3x-1)^{\frac{1}{2}}] ≥ 2x-3$

$\sqrt{(4x^2-3x-1)} ≥ 2x-3$

Determine the defined range

 $\sqrt{4x^2-3x-1} ≥ 2x-3 ,x\in[ ∞,-\frac{1}{4}] ∪ [1,+∞]$

Separate the inequality into 2 possible cases

$\sqrt{(4x^2-3x-1)} ≥ 2x-3 ,$ $2x-3 ≥ 0$

$\sqrt{(4x^2-3x-1)} ≥ 2x-3 ,$ $2x-3 <0$

Solve the inequality for x

$x≥\frac{10}{9},2x-3 ≥0$

$\sqrt{(4x^2-3x-1)} ≥ 2x-3 , 2x-3<0$

$x≥\frac{10}{9}, x≥\frac{3}{2}$

$\sqrt{(4x^2-3x-1)} ≥ 2x-3 , 2x-3<0$

Since the left hand side is always positive or zero, and the right hand side is always negative, the statement is true for any value of x

$x≥\frac{10}{9}, x≥\frac{3}{2}$

$x\in \reals, 2x-3<0$

$x\in \reals, x<\frac{3}{2}$

Find the intersection

$x\in [\frac{3}{2},+∞]$

$x\in \reals, x<\frac{3}{2}$

$x\in [-∞,\frac{3}{2}]$

Find the union

$x\in \reals , x\in[ -∞,-\frac{1}{4}] ∪ [1,+∞]$

Find the intersection of the defined solution and the defined range

$x\in[ -∞,-\frac{1}{4}] ∪ [1,+∞]$