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What range of values of x will satisfy the inequality |2x + [#permalink ]
25 May 2011, 06:04

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What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < -1/9 or x > 5

B. -1 < x < 1/9

C. -1/9 < x < 1

D. -1/9 < x < 5

E. x < -1/9 or x > 1

Last edited by

Bunuel on 28 Apr 2014, 01:06, edited 2 times in total.

Renamed the topic, edited the question and the OA.

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Re: What range of values of x will satisfy the inequality contd. [#permalink ]
25 May 2011, 06:26
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two methods here.

1. using the solutions here. x= -3/2 and x= 2/7 are solutions.

check for the regions

a. x<-3/2, b. -3/2<x<2/7 and c. x>2/7

values are a -(2x+3) > (7x-2) giving x <-1/9 hence not a solution.

b (2x+3) > (7x-2) giving x < 1 a solution.

c (2x+3) > -(7x-2) giving x>-1/9 hence the solution is -1/9 < x < 1

or

2. squaring both sides

(2x+3) ^2 > (7x-2)^2

gives 9x^2 -8x - 1 > 0 giving solution -1/9 < x < 1.

Hence C you can use whichever you are comfortable with.

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Re: What range of values of x will satisfy the inequality contd. [#permalink ]
25 May 2011, 14:42

Amit, Thank you, but could you please elaborate why in the option A "a. x<-3/2, values are a -(2x+3) > (7x-2) giving x <-1/9 hence not a solution." the second part - (7x-2) is positive? If we get a number less than -1/9, -10 for instance would not the second part be negative as well? Regards, Steve.

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Re: What range of values of x will satisfy the inequality contd. [#permalink ]
27 Apr 2014, 13:15

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Re: What range of values of x will satisfy the inequality |2x + [#permalink ]
28 Apr 2014, 01:06
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Re: What range of values of x will satisfy the inequality |2x + [#permalink ]
28 Apr 2014, 03:27

Bunuel wrote:

ramzin wrote:

What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|? A. x < -1/9 or x > 5 B. -1 < x < 1/9 C. -1/9 < x < 1 D. -1/9 < x < 5 E. x < -1/9 or x > 1

Since both sides of the inequality are non-negative we can safely square:

4x^2+12x+9>49x^2-28x+4 -->

45x^2-40x-5<0 -->

9x^2-8x-1<0 -->

(x+\frac{1}{9})(x-1)<0 .

The "roots" are -1/9 and 1 (Solving Quadratic Inequalities:

solving-quadratic-inequalities-graphic-approach-170528.html ). "<" sign indicates that the solution must be between the roots:

-\frac{1}{9}<x<1 .

Answer: C.

Hi Bunuel ,

Thanks for the answer. I did not understand how are the roots -1/9 and 1. I am getting the roots as 9 and 1. Can you please elaborate.

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Re: What range of values of x will satisfy the inequality |2x + [#permalink ]
28 Apr 2014, 03:38

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Re: What range of values of x will satisfy the inequality |2x + [#permalink ]
28 Apr 2014, 04:45

Bunuel wrote:

Bunuel wrote:

ramzin wrote:

What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|? A. x < -1/9 or x > 5 B. -1 < x < 1/9 C. -1/9 < x < 1 D. -1/9 < x < 5 E. x < -1/9 or x > 1

Since both sides of the inequality are non-negative we can safely square:

4x^2+12x+9>49x^2-28x+4 -->

45x^2-40x-5<0 -->

9x^2-8x-1<0 -->

(x+\frac{1}{9})(x-1)<0 .

The "roots" are -1/9 and 1 (Solving Quadratic Inequalities:

solving-quadratic-inequalities-graphic-approach-170528.html ). "<" sign indicates that the solution must be between the roots:

-\frac{1}{9}<x<1 .

Answer: C.

Hi Bunuel ,

Thanks for the answer. I did not understand how are the roots -1/9 and 1. I am getting the roots as 9 and 1. Can you please elaborate.

(x+\frac{1}{9})(x-1)=0 -->

x+\frac{1}{9}=0 or

x-1=0 -->

x=-\frac{1}{9} or

x=1 .[/quote]

That part is fine , i'm solving the quadratic equation 9x^2-8x-1 < 0 as (x-9)(x+1) then getting x = 9 , x = -1. I also check out your link for solving quadratic equations in equalities but could relate it.

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Re: What range of values of x will satisfy the inequality |2x + [#permalink ]
28 Apr 2014, 04:52
gauravsoni wrote:

Bunuel wrote:

Bunuel wrote:

What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < -1/9 or x > 5

B. -1 < x < 1/9

C. -1/9 < x < 1

D. -1/9 < x < 5

E. x < -1/9 or x > 1

Since both sides of the inequality are non-negative we can safely square:

4x^2+12x+9>49x^2-28x+4 -->

45x^2-40x-5<0 -->

9x^2-8x-1<0 -->

(x+\frac{1}{9})(x-1)<0 .

The "roots" are -1/9 and 1 (Solving Quadratic Inequalities:

solving-quadratic-inequalities-graphic-approach-170528.html ). "<" sign indicates that the solution must be between the roots:

-\frac{1}{9}<x<1 .

Answer: C.

Hi Bunuel ,

Thanks for the answer. I did not understand how are the roots -1/9 and 1. I am getting the roots as 9 and 1. Can you please elaborate.

(x+\frac{1}{9})(x-1)=0 -->

x+\frac{1}{9}=0 or

x-1=0 -->

x=-\frac{1}{9} or

x=1 .

That part is fine , i'm solving the quadratic equation 9x^2-8x-1 < 0 as (x-9)(x+1) then getting x = 9 , x = -1. I also check out your link for solving quadratic equations in equalities but could relate it.[/quote]

(x-9)(x+1) is not a correct factoring of 9x^2-8x-1, it should be

(x+\frac{1}{9})(x-1) (

(9x+1)(x-1) ).

Factoring Quadratics:

http://www.purplemath.com/modules/factquad.htm Solving Quadratic Equations:

http://www.purplemath.com/modules/solvquad.htm Hope this helps.

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Re: What range of values of x will satisfy the inequality |2x + [#permalink ]
29 Apr 2014, 00:49

amit2k9 wrote:

two methods here. 1. using the solutions here. x= -3/2 and x= 2/7 are solutions. check for the regions a. x<-3/2, b. -3/2<x<2/7 and c. x>2/7 values are a -( 2x+3) > (7x-2) giving x <-1/9 hence not a solution. b (2x+3) > (7x-2) giving x < 1 a solution. c (2x+3) > -( 7x-2) giving x>-1/9 hence the solution is -1/9 < x < 1 or 2. squaring both sides (2x+3) ^2 > (7x-2)^2 gives 9x^2 -8x - 1 > 0 giving solution -1/9 < x < 1. Hence C you can use whichever you are comfortable with.

Hi Bunuel,

Thanks for the solution! I get the squaring both sides approach, but it takes over 3 minutes for me to do it that way.

Is there a faster way to solve this?

I noticed Amit provided another method, but I'm not sure I understand the first approach completely. Could you explain why he used the "-" sign in the first and third parts (highlighted in blue above)?

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Re: What range of values of x will satisfy the inequality |2x + [#permalink ]
29 Apr 2014, 03:46
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gauravsoni wrote:

Bunuel wrote:

ramzin wrote:

What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|? A. x < -1/9 or x > 5 B. -1 < x < 1/9 C. -1/9 < x < 1 D. -1/9 < x < 5 E. x < -1/9 or x > 1

Since both sides of the inequality are non-negative we can safely square:

4x^2+12x+9>49x^2-28x+4 -->

45x^2-40x-5<0 -->

9x^2-8x-1<0 -->

(x+\frac{1}{9})(x-1)<0 .

The "roots" are -1/9 and 1 (Solving Quadratic Inequalities:

solving-quadratic-inequalities-graphic-approach-170528.html ). "<" sign indicates that the solution must be between the roots:

-\frac{1}{9}<x<1 .

Answer: C.

Hi Bunuel ,

Thanks for the answer. I did not understand how are the roots -1/9 and 1. I am getting the roots as 9 and 1. Can you please elaborate.

Hi Gauravsoni,

9x^2-8x-1<0 Can be factored 9x^2 -9x+x-1 <0

9x(x-1)+1 (x-1) <0 or (9x+1)(x-1)<0 or x=-1/9 or x=1

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Re: What range of values of x will satisfy the inequality |2x + [#permalink ]
07 May 2014, 09:41

hi bunnel

what do u mean when u say the following:

Quote:

Since both sides of the inequality are non-negative we can safely square: 4x^2+12x+9>49x^2-28x+4 --> 45x^2-40x-5<0 --> 9x^2-8x-1<0 --> (x+\frac{1}{9})(x-1)<0.

how how we solve inequality which is negative??

whenever we see absolute values on each side of the sign... should we use the squaring approach??

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Re: What range of values of x will satisfy the inequality |2x + [#permalink ]
08 May 2014, 02:24

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Re: What range of values of x will satisfy the inequality |2x + [#permalink ]
11 May 2014, 23:13

Bunuel wrote:

ramzin wrote:

What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|? A. x < -1/9 or x > 5 B. -1 < x < 1/9 C. -1/9 < x < 1 D. -1/9 < x < 5 E. x < -1/9 or x > 1

Since both sides of the inequality are non-negative we can safely square:

4x^2+12x+9>49x^2-28x+4 -->

45x^2-40x-5<0 -->

9x^2-8x-1<0 -->

(x+\frac{1}{9})(x-1)<0 .

The "roots" are -1/9 and 1 (Solving Quadratic Inequalities:

solving-quadratic-inequalities-graphic-approach-170528.html ). "<" sign indicates that the solution must be between the roots:

-\frac{1}{9}<x<1 .

Answer: C.

If we put in the value x = 0 , the inequality is not being satisfied , even though it is within the range. Am I missing something here ?..

I tried to deduce the answers by substituting possible values from the various option ranges.

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Re: What range of values of x will satisfy the inequality |2x + [#permalink ]
11 May 2014, 23:18
himanshujovi wrote:

Bunuel wrote:

ramzin wrote:

What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|? A. x < -1/9 or x > 5 B. -1 < x < 1/9 C. -1/9 < x < 1 D. -1/9 < x < 5 E. x < -1/9 or x > 1

Since both sides of the inequality are non-negative we can safely square:

4x^2+12x+9>49x^2-28x+4 -->

45x^2-40x-5<0 -->

9x^2-8x-1<0 -->

(x+\frac{1}{9})(x-1)<0 .

The "roots" are -1/9 and 1 (Solving Quadratic Inequalities:

solving-quadratic-inequalities-graphic-approach-170528.html ). "<" sign indicates that the solution must be between the roots:

-\frac{1}{9}<x<1 .

Answer: C.

If we put in the value x = 0 , the inequality is not being satisfied , even though it is within the range. Am I missing something here ?..

I tried to deduce the answers by substituting possible values from the various option ranges.

If x = 0, then |2x + 3| = 3 and |7x - 2| = 2 --> 3 > 2. So, 0 is a possible value of x.

Hope it helps.

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Re: What range of values of x will satisfy the inequality contd. [#permalink ]
18 Jun 2014, 05:15

amit2k9 wrote:

two methods here. 1. using the solutions here. x= -3/2 and x= 2/7 are solutions. check for the regions a. x<-3/2, b. -3/2<x<2/7 and c. x>2/7 values are a -(2x+3) > (7x-2) giving x <-1/9 hence not a solution. b (2x+3) > (7x-2) giving x < 1 a solution. c (2x+3) > -(7x-2) giving x>-1/9 hence the solution is -1/9 < x < 1 or 2. squaring both sides (2x+3) ^2 > (7x-2)^2 gives 9x^2 -8x - 1 > 0 giving solution -1/9 < x < 1. Hence C you can use whichever you are comfortable with.

Hi Amit, can you explain how to get the range -1/9 < x < 1 in the second method

the solutions are x< 1 and x < -1/9 (how does the sign reverse to the other side ?)

Re: What range of values of x will satisfy the inequality contd.
[#permalink ]
18 Jun 2014, 05:15