solve the inequality?

Question

squareroot 2(x-squareroot 2) < x+2 squareroot 2

Answer 1

sqrt(2) x - sqrt(2) < x + 2*sqrt(2)

First, group the terms with x in them on one side, and the ones without x on the right hand side.

sqrt(2)x - x < 2sqrt(2) + sqrt(2)

Factor the left hand side, group together the like terms on the right hand side.

x(sqrt(2) - 1) < 3sqrt(2)

Now, divide both sides by (sqrt(2) - 1)

x < 3sqrt(2) / [sqrt(2) - 1]

Let's change the right hand side around by rationalizing the denominator. We have to multiply top and bottom by [sqrt(2) + 1].

x < { [3sqrt(2)][sqrt(2) + 1] } / [sqrt(2) - 1][sqrt(2) + 1]
x < [3(2) + 3sqrt(2)] / (2 - 1)
x < 6 + 3sqrt(2)

Answer 2

2¹´²*(x - 2^¹/2) < x + 2*2¹´²
2¹´²x - 2 < x + 2*2¹´²
2¹´²x - x < 2 + 2*2¹´²
x(2¹´² - 1) < 2 + 2*2¹´²
x < (2 + 2*2¹´² )/(2¹´² - 1)
x < 2(1 + 2¹´² )/(2¹´² - 1)*(2¹´²+1)/(2¹´²+1)
x < 2(1 + 2¹´² )*(2¹´²+1)/(2 - 1)
x < 2(1 + 2¹´² )²
x < 2(1 + 2 + 2*2¹´²)
x < 2 + 4 + 4*2¹´²

x < 6 + 4*2¹´²

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Answer 3

0

Answer 4

√2(x-√2) < x+2√2
(√2)x - 2< x+2√2
(√2)x - x < (2√2) + 2
[(√2)-1] x < (2√2) + 2
x < [(2√2) + 2] / [(√2)-1]
x < 6+4√2
x < 11.657