Be careful if you do choose to multiply through by x; multiplying through by a negative number reverses an inequality and so if you choose this strategy you will have to split it up into two inequalities; one for x < 0 and one for x >= 0 and solve them independently.

A far better strategy is to sketch the graph of y = x + 4/x straight up if you can. If not multiply through by x^2 instead of x. This avoids the above issue because x^2 is never negative.

Reply 9

Star-girl

12

Original post by khanpatel321

x + 4/x ≥ 4

How do you solve this inequality.

You could compare the graphs of 4/x and 4-x.

Or you could do a case analysis:

First assume x is positive and multiply through. Solve that way and obtain a solution set for x>0.

Then assume that it is negative and multiply through, reverse the inequality and solve that way. Obtain a solution set for x<0.

Overall solution set is the union of the two sets.

x could be negative, which would reverse the inequality.

Reply 10

Kvothe the Arcane

20

Original post by Star-girl

You could compare the graphs of 4/x and 4-x.

Or you could do a case analysis:

First assume x is positive and multiply through. Solve that way and obtain a solution set for x>0.

Then assume that it is negative and multiply through, reverse the inequality and solve that way. Obtain a solution set for x<0.

Overall solution set is the union of the two sets.

x could be negative, which would reverse the inequality.

I would combine until you get

\dfrac{x^2-4x+4}{x} \geq 0

Multiply by x^2 as this is always positive to obtain

x(x^2-4x+4) \geq 0

and consider the regions when above the x axis.

Reply 11

Star-girl

12

1. Beware - the inequality in question includes equality.

2. This holds for all x>0 (x positive because that is what you assumed when you just multiplied through; can't have x=0 because we can't divide by zero and the inequality had 4/x in it), not just x>2.

3. x>2 and x<2 is impossible.

(edited 8 years ago)

Reply 12

khanpatel321

OP

19

Original post by 16Characters....

Be careful if you do choose to multiply through by x; multiplying through by a negative number reverses an inequality and so if you choose this strategy you will have to split it up into two inequalities; one for x < 0 and one for x >= 0 and solve them independently.

A far better strategy is to sketch the graph of y = x + 4/x straight up if you can. If not multiply through by x^2 instead of x. This avoids the above issue because x^2 is never negative.

Original post by Star-girl

You could compare the graphs of 4/x and 4-x.

Or you could do a case analysis:

First assume x is positive and multiply through. Solve that way and obtain a solution set for x>0.

Then assume that it is negative and multiply through, reverse the inequality and solve that way. Obtain a solution set for x<0.

Overall solution set is the union of the two sets.

x could be negative, which would reverse the inequality.

What is your final answer?

Reply 13

Star-girl

12

Original post by keromedic

I would combine until you get

\dfrac{x^2-4x+4}{x} \geq 0

Multiply by x^2 as this is always positive to obtain

x(x^2-4x+4) \geq 0

and consider the regions when above the x axis.

That works as well. :yep:

Reply 14

Star-girl

12

Original post by khanpatel321

What is your final answer?

From what we've suggested, what do you think the answer is rather than us just giving it to you?

Reply 15

khanpatel321

OP

19

Original post by Star-girl

From what we've suggested, what do you think the answer is rather than us just giving it to you?

if x >0

x^2 + 4 - 4x >= 0

x>=2

if x<0

-x^2 + 4x + 4 <= 0

Is this correct so far

Reply 16

Star-girl

12

Original post by khanpatel321

if x >0

x^2 + 4 - 4x >= 0

x>=2

if x<0

-x^2 + 4x + 4 <= 0

Is this correct so far

Bits in bold are wrong.

Reply 17

khanpatel321

OP

19

Original post by Star-girl

Bits in bold are wrong.

We need to multiply the both part of the inequality by x. When x <0, this will lead to change in the direction of the inequality. But for x < 04/x < 0, too, so in that case x +4/x< 0and x < cannot be a solution of the original inequality.

So if x is a solution then x > 0 and we can multiply the both parts of the inequality by x,obtaining

x^2 + 4 ≥ 4
x^2 − 4x + 4 ≥ 0
(x − 2)^2 ≥ 0

But the last inequality holds for all real x. Hence our solutions are bound only by restriction made in the process of rearrangements,x > 0, and therefore the solution set is ]0, +∞[

Is this how you worked it out ?

Reply 18

Star-girl

12

Original post by khanpatel321

We need to multiply the both part of the inequality by x. When x <0, this will lead to change in the direction of the inequality. But for x < 04/x < 0, too, so in that case x +4/x< 0and x < cannot be a solution of the original inequality.

So if x is a solution then x > 0 and we can multiply the both parts of the inequality by x,obtaining

x^2 + 4 ≥ 4
x^2 − 4x + 4 ≥ 0
(x − 2)^2 ≥ 0

But the last inequality holds for all real x. Hence our solutions are bound only by restriction made in the process of rearrangements,x > 0, and therefore the solution set is ]0, +∞[

Is this how you worked it out ?

Similar in parts, yes.