Task 1: abs(7-3x) + 4x > 5

Task 2: sqrt(7-3x) <= x -1

Task 3: sqrt(5-2x) > 1 - x

Task 4: x^2*(1 + 4x^2) < abs(x - 2x^2) + 6 + 4x^3

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- Oct 25th 2012, 08:10 AMmiloserdowSolve sevral simple inequalities with modulus and sqrts.
Task 1: abs(7-3x) + 4x > 5

Task 2: sqrt(7-3x) <= x -1

Task 3: sqrt(5-2x) > 1 - x

Task 4: x^2*(1 + 4x^2) < abs(x - 2x^2) + 6 + 4x^3 - Oct 25th 2012, 09:03 AMHallsofIvyRe: Solve sevral simple inequalities with modulus and sqrts.
If 7- 3x< 0 (that is, if x> 7/3) this is 3x- 7+ 4x> 5. Can you solve that? If $\displaystyle x\ge 7/3$ so that $\displaystyle 7x- 3\ge 0$, this is $\displaystyle 7- 3x+ 4x> 5$. Can you solve that?

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Task 2: sqrt(7-3x) <= x -1

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]Task 3: sqrt(5-2x) > 1 - x

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Task 4: x^2*(1 + 4x^2) < abs(x - 2x^2) + 6 + 4x^3

That is, there are two cases:

1) x< 0 or x> 1/2. In this case, $\displaystyle x- 2x^2< 0$ so $\displaystyle |x- 2x^2|= 2x^2- x$ and the inequality becomes $\displaystyle x^2*(1 + 4x^2) < 2x^2- x + 6 + 4x^3$ which is the same as $\displaystyle 4x^4+ x^2< 4x^3+ 2x^2- x+ 6$ which is itself the same as $\displaystyle 4x^4- 4x^3+ x^2+ x- 6< 0$. The simplest way to solve that is to find the solutions to $\displaystyle 4x^4- 4x^3+ x^2+ x- 6= 0$ and check points between those solutions to see where they are "> 0" and where "< 0".

2) 0< x< 1/2. In this case, $\displaystyle x- 2x^2> 0$ so $\displaystyle |x- 2x^2|= x- 2x^2$ and the inequality becomes $\displaystyle x^2*(1 + 4x^2) < x- 2x^2 + 6 + 4x^3$ which is the same as $\displaystyle 4x^4+ x^2< 4x^3- 2x^2+ x+ 6$ which is itself the same as $\displaystyle 4x^4- 4x^3+ 3x^2- x- 6< 0$. - Oct 25th 2012, 03:57 PMmiloserdowRe: Solve sevral simple inequalities with modulus and sqrts.
Thank you!