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Re: What range of values of x will satisfy the inequality contd. [#permalink]

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27 Apr 2014, 14:15

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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.

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

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)?

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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Hope to clear it this time!! GMAT 1: 540 Preparing again

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??

We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality)

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 |2x + [#permalink]

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23 Jul 2015, 14:22

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Re: What range of values of x will satisfy the inequality |2x + [#permalink]

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28 Mar 2016, 08:03

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

Plotting the The halfs of the equation we get the diagram as shown...shaded portion shows the area of interest satisfyin the above inequality. In upper half x=1 is the point of intersection i.e the maximum value of x. Only C satisfies.

Re: What range of values of x will satisfy the inequality |2x + [#permalink]

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10 Apr 2016, 19:27

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

I took the safest approach, plug in numbers :D

A. x=-2. -4+3 = -1, |-1|=1. 7*-2 = -14 -2 = -16. |-16|=16 2 is not > than 16, so can't be A. B. x= -1/8. -2/8+3 = 2.75. 7*-1/8 = -7/8 -2 = |-2.875| not true, so can't be B. D. x=4. 8+3=11. 28-2=26. so not true, D is out. E. same as in A. x=-2 doesn't work.

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