Answer to Question #223005 in Linear Algebra for Vanessa

Question

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

Answer 1

$\sqrt{x^2-2x-8} ≤ -x+2$

$\sqrt{x^2-2x-8} ≤ -x+2$ $,x∈[∞,− 2]∪[4,+∞]$

Separate into possible cases

$\sqrt{x^2-2x-8} ≤ -x+2, -x+2≥0$

$\sqrt{x^2-2x-8} ≤ -x+2, -x+2<0$

Solve for inequality x

$x≤6,-x+2≥0$

$\sqrt{x^2-2x-8} ≤ -x+2, -x+2<0$

$x≤6,x≤2$

$\sqrt{x^2-2x-8} ≤ -x+2, -x+2<0$

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

$x≤6,x≤2$

$x\in\varnothing,-1+2<0$

$x\in\varnothing,1>2$

Find the intersection

$x\in [-∞,-2]$

$x\in\varnothing$

Find the union

$x∈[-∞,2],x∈[-∞,− 2]∪[4,+∞]$

Find the intersection of the solution and the defined range

$x\in [-∞,-2]$

Alternate form

$x≤-2$