Solve sevral simple inequalities with modulus and sqrts.

**miloserdow** · October 25th 2012, 08:10 AM

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

**HallsofIvy** · October 25th 2012, 09:03 AM

Originally Posted by miloserdow

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

If 7- 3x< 0 (that is, if x> 7/3) this is 3x- 7+ 4x> 5. Can you solve that? If $x\ge 7/3$ so that $7x- 3\ge 0$ , this is $7- 3x+ 4x> 5$ . Can you solve that?

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

$\sqrt{7- 3x}$ is positive and if a< b, then $a^2< b^2$ so this is the same as [itex]7- 3x\le (x- 1)^2[/itex]. Can you solve that?

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

Same thing as task 2.

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

As in task 1, consider the cases $x- 2x^2\ge 0$ and $x- 2x^2< 0$ separately. The equation $x- 2x^2= x(1- 2x)= 0$ has solutions x= 0 and x= 1/2. If x= -1, $x- 2x^2= -1- 2= -3< 0$ so $x- 2x^2< 0$ for all x < 0. If x= 1/4, then $x- 2x^2= 1/2- 2/16= 4/8- 1/8= 3/8> 0$ so $x- 2x^2> 0$ for all x between 0 and 1/2. Finally, if x= 2, $x- 2x^2= 2- 8= -6< 0$ so that $x- 2x^2< 0$ for all x> 1.

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

2) 0< x< 1/2. In this case, $x- 2x^2> 0$ so $|x- 2x^2|= x- 2x^2$ and the inequality becomes $x^2*(1 + 4x^2) < x- 2x^2 + 6 + 4x^3$ which is the same as $4x^4+ x^2< 4x^3- 2x^2+ x+ 6$ which is itself the same as $4x^4- 4x^3+ 3x^2- x- 6< 0$ .

**miloserdow** · October 25th 2012, 03:57 PM

Thank you!

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