Answer to Question #223005 in Linear Algebra for Vanessa

Question #223005


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


1
Expert's answer
2021-08-22T18:14:12-0400

"\\sqrt{x^2-2x-8} \u2264 -x+2"



"\\sqrt{x^2-2x-8} \u2264 -x+2" ",x\u2208[\u221e,\u2212 2]\u222a[4,+\u221e]"


Separate into possible cases


"\\sqrt{x^2-2x-8} \u2264 -x+2, -x+2\u22650"


"\\sqrt{x^2-2x-8} \u2264 -x+2, -x+2<0"


Solve for inequality x


"x\u22646,-x+2\u22650"


"\\sqrt{x^2-2x-8} \u2264 -x+2, -x+2<0"


"x\u22646,x\u22642"


"\\sqrt{x^2-2x-8} \u2264 -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\u22646,x\u22642"

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

"x\\in\\varnothing,1>2"

Find the intersection

"x\\in [-\u221e,-2]"

"x\\in\\varnothing"

Find the union

"x\u2208[-\u221e,2],x\u2208[-\u221e,\u2212 2]\u222a[4,+\u221e]"


Find the intersection of the solution and the defined range


"x\\in [-\u221e,-2]"


Alternate form

"x\u2264-2"





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