Inequalities and Roots

Question

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.

KarishmaB · Accepted Answer

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

KarishmaB · Answer

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

GyanOne · Answer

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 [ 20.1 KiB | Viewed 24516 times ]

valentindima · Answer

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.

Asher · Answer

Thanks GyanOne for the explanation and the graph. I get it now.

Asher · Answer

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.

Asher · Answer

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)

krishp84 · Answer

√(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

Asher · Answer

Thanks krishp84.

KarishmaB · Answer

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.

Asher · Answer

Now i get it.

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.

sdas · Answer

great explanations all. i got some good insights in an area i hv been struggling on (apart from sequences :cry:

)

kartik222 · Answer

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,

keenys · Answer

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?

KarishmaB · Answer

\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.

gmatter0913 · Answer

Could somebody help me with the below problems?

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

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

KarishmaB · Answer

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

gmatter0913 · Answer

In one of your earlier posts, you mentioned

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.

Could you help me understand this please?

gmatter0913 · Answer

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?

Inequalities and Roots

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