# Quadratic Inequalities

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Thread starter 5 years ago
#1
x + 4/x ≥ 4

How would you solve this inequality
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5 years ago
#2
multiply through by x, take all to one side, factorise, solve for x, draw graph, evaluate < or >
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Thread starter 5 years ago
#3
(Original post by pippabethan)
multiply through by x, take all to one side, factorise, solve for x, draw graph, evaluate < or >
Doesn't work
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5 years ago
#4
4 ≥ 3x solve from here.
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5 years ago
#5
it does. x^2 -4x +4 > 0
(x-2)(x-2)>0
x=2 (one root of eq)
so x =/= 2 but x>2 and x<2
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5 years ago
#6
(Original post by Econhope20)
4 ≥ 3x solve from here.
Where did that come from?
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5 years ago
#7
(Original post by khanpatel321)
Doesn't work

it does. x^2 -4x +4 > 0
(x-2)(x-2)>0
x=2 (one root of eq)
so x =/= 2 but x>2 and x<2
0
reply
Thread starter 5 years ago
#8
(Original post by pippabethan)
it does. x^2 -4x +4 > 0
(x-2)(x-2)>0
x=2 (one root of eq)
so x =/= 2 but x>2 and x<2
So what is the answer?
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5 years ago
#9
(Original post by khanpatel321)
So what is the answer?
yes. more importantly, do you see how?
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5 years ago
#10
(Original post by pippabethan)
Where did that come from?
multiplying both sides by x and moving the x from x+4 to the other side, your working is wrong i think.
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Thread starter 5 years ago
#11
(Original post by pippabethan)
yes. more importantly, do you see how?
But that is not the answer....
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5 years ago
#12
(Original post by Econhope20)
multiplying both sides by x and moving the x from x+4 to the other side, your working is wrong i think.
it's not. you've forgotten to multiply x by x
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5 years ago
#13
(Original post by khanpatel321)
But that is not the answer....
is the whole of the LHS divided by x?
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5 years ago
#14
(Original post by pippabethan)
it's not. you've forgotten to multiply x by x
is the question (x+4)/x or x+4/x? OP hasnt made that clear
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5 years ago
#15
(Original post by khanpatel321)
So what is the answer?
x = 2
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5 years ago
#16
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.
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5 years ago
#17
(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.

(Original post by pippabethan)
multiply through by x, take all to one side, factorise, solve for x, draw graph, evaluate < or >
x could be negative, which would reverse the inequality.
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5 years ago
#18
(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

Multiply by x^2 as this is always positive to obtain and consider the regions when above the x axis.
1
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5 years ago
#19
(Original post by pippabethan)
it does. x^2 -4x +4 > 0
(x-2)(x-2)>0
x=2 (one root of eq)
so x =/= 2 but x>2 and x<2
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.
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Thread starter 5 years ago
#20
(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?
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