# Quadratic Inequalities

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#2

multiply through by x, take all to one side, factorise, solve for x, draw graph, evaluate < or >

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(Original post by

multiply through by x, take all to one side, factorise, solve for x, draw graph, evaluate < or >

**pippabethan**)multiply through by x, take all to one side, factorise, solve for x, draw graph, evaluate < or >

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#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

(x-2)(x-2)>0

x=2 (one root of eq)

so x =/= 2 but x>2 and x<2

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#7

(Original post by

Doesn't work

**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

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(Original post by

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

**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

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#9

(Original post by

So what is the answer?

**khanpatel321**)So what is the answer?

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#10

(Original post by

Where did that come from?

**pippabethan**)Where did that come from?

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(Original post by

yes. more importantly, do you see how?

**pippabethan**)yes. more importantly, do you see how?

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#12

(Original post by

multiplying both sides by x and moving the x from x+4 to the other side, your working is wrong i think.

**Econhope20**)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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#13

(Original post by

But that is not the answer....

**khanpatel321**)But that is not the answer....

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#14

(Original post by

it's not. you've forgotten to multiply x by x

**pippabethan**)it's not. you've forgotten to multiply x by x

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#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.

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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#17

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.

**pippabethan**)

multiply through by x, take all to one side, factorise, solve for x, draw graph, evaluate < or >

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#18

(Original post by

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.

**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.

Multiply by x^2 as this is always positive to obtain and consider the regions when above the x axis.

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#19

**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

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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(Original post by

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.

**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.

**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.

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