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Inequalities and Roots

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Asher
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Inequalities and Roots [#permalink] New post   09 Aug 2011, 12:56
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Something interesting that i read while searching for material on how to solve inequalities with roots.

Though i would share it and also clarify a few doubts. I have highlighted the portions that were confusing, in blue

Kind of backbone for solving inequalities with roots, √x>y OR √x<y

• √x is undefined if x<0
• both sides can be squared when x≥0 and y≥0
• if √x>y is identically true if √x≥0 and y<0 what does identically true/false mean
• But √x<y is identically false if y<0



e.g., √(2x+3) > x

when, √(2x+3) ≥ 0 and x<0

left side, 2x + 3 ≥ 0 => x ≥ -1.5

right side, x < 0

thus, -1.5 ≤ x <0 (partial solution)

2nd Condition, where both left and right side, ≥ 0

(√(2x+3))^2 > x^2
=> 2x + 3 > x^2
=> x^2 -2x -3 <0
=> (x+1) (x-3)
=> x = -1 or 3

Since the graph is a parabola, it attains its negative values at -1<x<3
since x ≥ 0, thus 0 ≤ x < 3 (partial solution)

So the inequality becomes, -1.5 ≤ x <3

could some show me how the parabola would look like.

BTW, here are a few more inequalities, if someone wants to try out:

1. √(3x-2) < 2x-3
2. √(2x - 5) > -4x + 3

please share your thoughts on this.
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Re: Inequalities and Roots [#permalink] New post   09 Aug 2011, 21:05
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Let's try one of the questions you have given.

1. √(3x-2) < 2x-3
Left hand side is positive, we know. Also, the term under the root i.e. (3x - 2) should be positive or 0. 3x - 2>= 0 i.e. x >= 2/3.

What about the right hand side? Here we can say that the right hand side will definitely be positive too since left hand side (a positive number (√3x-2)) is less than the right hand side. Hence, 2x - 3 > 0
x > 3/2

Let's square both sides now:
3x - 2 < (2x - 3)^2
4x^2 -15x + 11 > 0
4(x - 1)(x - 11/4) > 0
Since the right most region is positive, we will get:
positive ... 1 ... negative ... 11/4 ... positive

So, 1 > x or x > 11/4.

But we saw above that x > 3/2
So x > 11/4
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Re: Inequalities and Roots [#permalink] New post   09 Aug 2011, 21:03
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Asher wrote:

• √x is undefined if x<0
• both sides can be squared when x≥0 and y≥0
• if √x>y is identically true if √x≥0 and y<0 what does identically true/false mean
• But √x<y is identically false if y<0




Some Explanations:
1. √x implies the positive square root of x. So √x is positive. But we do not know whether the right side of the inequality is positive or negative. Hence, we cannot square the inequality.
Note: Only when both sides are positive, you can square the inequality and still retain the same relation. e.g.
2 < 3
2^2 < 3^2

But
-2 < 1
4 not less than 1

Similarly
-4 < -2
16 not less than 4

So before you square both sides of the inequality, always ensure that both sides are positive

2. √x>y
We know that √x is positive. If y is negative, then this inequality will always hold since positive > negative.

3. √x<y
We know that √x is positive. If y is negative, then this inequality will never hold since positive is never less than negative.
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Re: Inequalities and Roots [#permalink] New post   11 Aug 2011, 23:22
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Asher wrote:
Karishma, here's me trying to solve the second problem. Let me know if i got it right.

2. √(2x - 5) > -4x + 3

Since sq.root is always positive, thus:

√(2x - 5) ≥ 0
2x - 5 ≥ 0
x ≥ 5/2


No problems till here... we know now that x must be > or equal to 5/2.

Next think, we have LHS > RHS. RHS can be negative or positive. So there are two cases possible:
Case 1: -4x + 3 < 0
x > 3/4 ----- (1)
Since in this case, RHS is negative and LHS will always be positive, LHS will be > than RHS. So the inequality will hold whenever x > 3/4 and x >= 5/2.
So if x >= 5/2, the inequality will hold.

Case 2: -4x + 3 > 0
x < 3/4 ------(2)
Now think, is it possible that x is < than 3/4 and greater than 5/2? No! So this case doesn't give any solutions.

Hence, the only solution is x >= 5/2

and yes, don't try to learn up Mathematical concepts. Try to understand them. That way, you will never forget them.
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Re: Inequalities and Roots [#permalink] New post   12 Mar 2013, 21:08
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keenys wrote:
sqroot(11-5x)>x-1



\(\sqrt{11 - 5x} > x - 1\)

Terms inside square roots are never negative so,
11 - 5x >= 0
x <= 11/5
x <= 2.2

You need to take two cases because the right hand side can be positive or negative.

If (x - 1) < 0
x < 1
In that case, the inequality will always hold because left hand side will be non-negative.

If (x - 1) >= 0, x >= 1
Square both sides of \(\sqrt{11 - 5x} > x - 1\)
11 - 5x > x^2 + 1 - 2x
x^2 + 3x - 10 < 0
(x + 5) (x - 2) < 0
-5 < x < 2
Since x >= 1, 1 <= x < 2

So, either x should be less than 1 or between 1 (inclusive) and 2. Hence, whenever x < 2, the inequality will hold.
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Re: Inequalities and Roots [#permalink] New post   Updated on: 09 Aug 2011, 21:08
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Identically true means true for all values of the variable.

For example:
sqrt(x)>y if sqrt(x)>=0 and y<0 means that this relation holds for all values of x and y that satisfy sqrt(x)>=0 and y<0.

The graph of the parabola is attached.

Please note:
The graph of ax^2+bx+c:
(a) Is a parabola always
(b) Opens upwards (towards +y axis) if a>0 and downwards (towards -y axis) if a<0
(c) Intersects the x axis at the roots of the equation ax^2 + bx + c = 0
(d) Will not touch the x axis if the roots are not real (if b^2 - 4ac < 0)
(e) Has its lowest point at x= -b/2a
Attachments

parabola.jpg
parabola.jpg [ 20.1 KiB | Viewed 21998 times ]


Originally posted by GyanOne on 09 Aug 2011, 13:22.
Last edited by GyanOne on 09 Aug 2011, 21:08, edited 1 time in total.
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Re: Inequalities and Roots [#permalink] New post   23 Jul 2013, 22:26
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gmatter0913 wrote:
Could somebody help me with the below problems?

1. x+2 < sqrt (x+14)

2. x-1 < sqrt(7-x)


Use the concepts given above to try to solve these:

\(x+2 < \sqrt{(x+14)}\)

The quantity under the root must be non negative so x >= -14

Now left hand side i.e. x+2 can be positive, 0 or negative. Take two cases:

Case 1: x + 2 >= 0
x >= -2
Now both sides of the inequality are non negative so we can square it:
\((x + 2)^2 < (x + 14)\)
\(x^2 + 3x - 10 < 0\)
\((x + 5) (x - 2) < 0\)
\(-5 < x < 2\)

Since x >= -2, we get -2 <= x < 2

Case 2: x + 2 < 0
x < -2
The left hand side i.e. x + 2 is always negative in this case while right hand side is always non negative so this inequality will hold for all values in this range.
-14 <= x < -2

So the overall acceptable range is -14<= x < 2
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Re: Inequalities and Roots [#permalink] New post   10 Aug 2011, 17:45
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Asher wrote:
Karishma, here's me trying to solve the second problem. Let me know if i got it right.

2. √(2x - 5) > -4x + 3

Since sq.root is always positive, thus:

√(2x - 5) ≥ 0
2x - 5 ≥ 0
x ≥ 5/2

If, -4x + 3 < 0
x > 3/4 ----- (1)

If, -4x + 3 > 0
x < 3/4 ------(2)

How can the left side be both x> 3/4 and x< 3/4

For, -4x + 3 > 0 we get x < 3/4 (can we get rid of this ans right away since LHS with sq. root is x ≥ 5/2 ),

and only compare RHS x > 3/4 and LHS x ≥ 5/2 to conclude that the ans. is x ≥ 5/2

Anyways, If we square both side when both sides for ≥ 0, we get no real roots.

(√(2x - 5))^2 > (-4x + 3)^2
2x - 5 > 16x^2 - 24x + 9
0> 16x^2 - 26x + 14
0> 8x^2 - 13x + 7
Here b^2 - 4ac = 169 - 224 = - 55 (no roots)


√(2x - 5) > -4x + 3

Since sq.root is always positive, thus:

√(2x - 5) ≥ 0
2x - 5 ≥ 0
x ≥ 5/2 ------------------------------(1)

If x ≥ 5/2 then -4x <= -10
==> -4x+3 <= -7 ---------------------(2)

Now as per your question -
-4x + 3 is less than 0...This is the super set of (2), so there is no other unique solution...
other than x ≥ 5/2
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Re: Inequalities and Roots [#permalink] New post   25 Jul 2013, 21:26
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gmatter0913 wrote:
Hi Karishma,

I tried the problem x-1 < sqrt (7-x) as below:

As 7-x is under sqrt, it is +ve. Therefore, 7-x>=0 ; x<=7 ----->(1)

x-1 can be -ve or +ve

When x-1<=0; x<=1 --------> (2)

When x-1>=0; x>=1 --------> (3)

As both sides are +ve, we can square both the sides

(x-1)^2 < 7-x
x^2 -x -6<0
(x-3)(x+2)<0

-2<x<3 ------------>(4)

The answer to this problem is (x<3). I am not sure how to arrive at that from hereon. Could you please help me?


You have done the process correctly. Now you need to understand what this implies.

You got x <= 7

Case 1: x-1< 0
When x-1< 0; x < 1
Note that when x-1 is negative, it will always be less than \(\sqrt{(7-x)}\)
So whenever x<1, the inequality will always hold.

Case 2: x-1 >= 0
If x-1 is non negative, we can square the inequality.
From this, you get -2<x<3.
The inequality holds in this range.

From the two cases, we see that the inequality holds for the range x < 3.
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Re: Inequalities and Roots [#permalink] New post   26 Jul 2013, 04:23
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gmatter0913 wrote:
Hi Karishma,

I have one more doubt on my solution posted earlier.

Quote:
I tried the problem x-1 < sqrt (7-x) as below:

As 7-x is under sqrt, it is +ve. Therefore, 7-x>=0 ; x<=7 ----->(1)

x-1 can be -ve or +ve

When x-1<=0; x<=1 --------> (2)

When x-1>=0; x>=1 --------> (3)

As both sides are +ve, we can square both the sides

(x-1)^2 < 7-x
x^2 -x -6<0
(x-3)(x+2)<0

-2<x<3 ------------>(4) Shouldn't this be 1<=x<3 (as x>=1 is the pre-supposed condition to square them)


Most certainly. The only reason I don't care about the values from -2 to 1 is that these values are already covered in the first case. We know they already hold for the inequality. We only get 1 to 3 extra values and that's what we care about.

In a stand alone question, the presupposed condition must be satisfied (x <= 7 AND x >= 1)
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Re: Inequalities and Roots [#permalink] New post   09 Aug 2011, 13:52
Nice.

I have forgotten how to solve these. Thanks for reminding.

You can visualize very well the solution to this by plotting the graph. Not that you can do it during the actual exam, but for reference and to remember how the typical functions look like you can use //rechneronline.de/function-graphs/ to graph any function.
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Re: Inequalities and Roots [#permalink] New post   10 Aug 2011, 02:29
GyanOne wrote:
Identically true means true for all values of the variable.

For example:
sqrt(x)>y if sqrt(x)>=0 and y<0 means that this relation holds for all values of x and y that satisfy sqrt(x)>=0 and y<0.

The graph of the parabola is attached.

Please note:
The graph of ax^2+bx+c:
(a) Is a parabola always
(b) Opens upwards (towards +y axis) if a>0 and downwards (towards -y axis) if a<0
(c) Intersects the x axis at the roots of the equation ax^2 + bx + c = 0
(d) Will not touch the x axis if the roots are not real (if b^2 - 4ac < 0)
(e) Has its lowest point at x= -b/2a


Thanks GyanOne for the explanation and the graph. I get it now.
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Re: Inequalities and Roots [#permalink] New post   10 Aug 2011, 02:38
VeritasPrepKarishma wrote:
Asher wrote:

• √x is undefined if x<0
• both sides can be squared when x≥0 and y≥0
• if √x>y is identically true if √x≥0 and y<0 what does identically true/false mean
• But √x<y is identically false if y<0




Some Explanations:
1. √x implies the positive square root of x. So √x is positive. But we do not know whether the right side of the inequality is positive or negative. Hence, we cannot square the inequality.
Note: Only when both sides are positive, you can square the inequality and still retain the same relation. e.g.
2 < 3
2^2 < 3^2

But
-2 < 1
4 not less than 1

Similarly
-4 < -2
16 not less than 4

So before you square both sides of the inequality, always ensure that both sides are positive

2. √x>y
We know that √x is positive. If y is negative, then this inequality will always hold since positive > negative.

3. √x<y
We know that √x is positive. If y is negative, then this inequality will never hold since positive is never less than negative.


Thanks karishma for the explanation.

i now realize that when i read the material found on one of the math websites, i kind of just tried to mug the concept without really understanding it. By now the concept it clear.
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Re: Inequalities and Roots [#permalink] New post   10 Aug 2011, 03:33
Karishma, here's me trying to solve the second problem. Let me know if i got it right.

2. √(2x - 5) > -4x + 3

Since sq.root is always positive, thus:

√(2x - 5) ≥ 0
2x - 5 ≥ 0
x ≥ 5/2

If, -4x + 3 < 0
x > 3/4 ----- (1)

If, -4x + 3 > 0
x < 3/4 ------(2)

How can the left side be both x> 3/4 and x< 3/4

For, -4x + 3 > 0 we get x < 3/4 (can we get rid of this ans right away since LHS with sq. root is x ≥ 5/2 ),

and only compare RHS x > 3/4 and LHS x ≥ 5/2 to conclude that the ans. is x ≥ 5/2

Anyways, If we square both side when both sides for ≥ 0, we get no real roots.

(√(2x - 5))^2 > (-4x + 3)^2
2x - 5 > 16x^2 - 24x + 9
0> 16x^2 - 26x + 14
0> 8x^2 - 13x + 7
Here b^2 - 4ac = 169 - 224 = - 55 (no roots)
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Re: Inequalities and Roots [#permalink] New post   11 Aug 2011, 20:15
Quote:
Now as per your question -
-4x + 3 is less than 0...This is the super set of (2), so there is no other unique solution...
other than x ≥ 5/2


Thanks krishp84.
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Re: Inequalities and Roots [#permalink] New post   13 Aug 2011, 03:18
Now i get it. :)

Quote:
and yes, don't try to learn up Mathematical concepts. Try to understand them. That way, you will never forget them.


Thanks for this advice. This makes soo much sense. Earlier i would just learn how to solve one particular problem without really understanding the concept, but then i would always make a mistake on a different problem based on the same concept.

Anyways, my POA now is to brush my basic quant skills.
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Re: Inequalities and Roots [#permalink] New post   15 Jan 2012, 05:28
great explanations all. i got some good insights in an area i hv been struggling on (apart from sequences :cry: )
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Re: Inequalities and Roots [#permalink] New post   30 Apr 2012, 01:19
hi All,

:? Might sound lame but what is the question being asked here:

Let's try one of the questions you have given.

1. √(3x-2) < 2x-3

Whats the question here? what are we trying to solve for? Is a P.S question or D.s question?

thanks,
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Re: Inequalities and Roots [#permalink] New post   12 Mar 2013, 09:58
Hi Karishma,

Thanks for the great explanation. However, I am stuck in solving the below question:

sqroot(11-5x)>x-1

Could you please help?
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Re: Inequalities and Roots [#permalink] New post   23 Jul 2013, 05:54
Could somebody help me with the below problems?

1. x+2 < sqrt (x+14)

2. x-1 < sqrt(7-x)
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