Question

Solve the inequality √(4x²-8x+7) ≤ -2x

Answer 1

$\sqrt{4x^2-8x+7}\leq-2x$

$\sqrt{4x^2-8x+7} ≤ -2x, x\isin\reals$

Separate into two possible case

$\sqrt{4x^2-8x+7}≤ -2x,-2x≥0$

$\sqrt{4x^2-8x+7}≤ -2x,-2x<0$

Solve the inequality for x

$x≥\frac{7}{8}, -2x≥0$

$\sqrt{4x^2-8x+7}≤ -2x,-2x<0$

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

$x≥\frac{7}{8},x≤0$

$\sqrt{4x^2-8x+7}≤ -2x,-2x<0$

$x\isin\varnothing$ , $-2x<0$

$x\isin\varnothing,x>0$

Find the intersection

$x\isin\varnothing$

$x\isin\varnothing,x>0$

The union

$x\isin\varnothing$

That is, there is no solution.