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New Algebra Set!!! [#permalink]
18 Mar 2013, 06:56

27

This post received KUDOS

Expert's post

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The next set of medium/hard PS algebra questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers.

1. If x=\sqrt[4]{x^3+6x^2}, then the sum of all possible solutions for x is:

Re: New Algebra Set!!! [#permalink]
22 May 2013, 21:19

Bunuel wrote:

[ Take the given expression to the 4th power: x^4=x^3+6x^2;

Re-arrange and factor out x^2: x^2(x^2-x-6)=0;

Factorize: x^2(x-3)(x+2)=0;

So, the roots are x=0, x=3 and x=-2. But x cannot be negative as it equals to the even (4th) root of some expression (\sqrt{expression}\geq{0}), thus only two solution are valid x=0 and x=3.

The sum of all possible solutions for x is 0+3=3.

Answer: D.

Hi Bunuel, Request you to please let me know where I'm making a mistake.

Since x^3+6*x^2 >= 0 Then, x^2(x+6)>=0

i.e (x-0)^2(x-(-6))>=0 This implies thatx>=-6 Hence, x=-2 is a valid root, and sum of all roots should bex=3+(-2) = 1 Please let me know where I am going wrong.

Re: New Algebra Set!!! [#permalink]
23 May 2013, 01:00

Expert's post

imhimanshu wrote:

Bunuel wrote:

[ Take the given expression to the 4th power: x^4=x^3+6x^2;

Re-arrange and factor out x^2: x^2(x^2-x-6)=0;

Factorize: x^2(x-3)(x+2)=0;

So, the roots are x=0, x=3 and x=-2. But x cannot be negative as it equals to the even (4th) root of some expression (\sqrt{expression}\geq{0}), thus only two solution are valid x=0 and x=3.

The sum of all possible solutions for x is 0+3=3.

Answer: D.

Hi Bunuel, Request you to please let me know where I'm making a mistake.

Since x^3+6*x^2 >= 0 Then, x^2(x+6)>=0

i.e (x-0)^2(x-(-6))>=0 This implies thatx>=-6 Hence, x=-2 is a valid root, and sum of all roots should bex=3+(-2) = 1 Please let me know where I am going wrong.

Thanks

Plug x=-2 into x=\sqrt[4]{x^3+6x^2}. Does the equation hold true? _________________

Re: New Algebra Set!!! [#permalink]
23 May 2013, 04:03

2. The equation x^2 + ax - b = 0 has equal roots, and one of the roots of the equation x^2 + ax + 15 = 0 is 3. What is the value of b?

A. -64 B. -16 C. -15 D. -1/16 E. -1/64

Solution:

x^2 + ax + 15 = 0 equation has one root as 3. only possible values of getting product of the roots 15 , having one root 3 is 5*3. x^2-5x-3x+15=0 take common values out => x(x-5)-3(x-5) =0 => x=5 and x=3(what was given) , from this we get a= -8.

x^2 -8x - b =0 => if roots are equal then b^2-4ac=0, 64+4b=0 from this b=-16

Re: New Algebra Set!!! [#permalink]
23 Jun 2013, 00:37

Expert's post

aquax wrote:

Hi,

For the first question, what do you mean by "But X cannot be negative as it equals to the even (4th) root of some expression"?

x cannot be negative as it equals to the even (4th) root of some expression (\sqrt[even]{expression}\geq{0}), thus only two solution are valid x=0 and x=3.

When the GMAT provides the square root sign for an even root, such as \sqrt{x} or \sqrt[4]{x}, then the only accepted answer is the positive root. That is, \sqrt{25}=5, NOT +5 or -5. Even roots have only non-negative value on the GMAT.

In contrast, the equation x^2=25 has TWO solutions, \sqrt{25}=+5 and -\sqrt{25}=-5.

So, we need to maximize the value of -3(x-2)^2-2(y+3)^2-9.

Since, the maximum value of -3(x-2)^2 and -2(y+3)^2 is zero, then the maximum value of the whole expression is 0+0-9=-9.

Answer: B.

Hi Bunuel,

Just as a matter of fact can you please explain that how you reached to factorization. In other words how a candidate will decide how to split -39 between (X,Y).

Your reply is appreciated !!

Rgds, TGC !! _________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

Re: New Algebra Set!!! [#permalink]
24 Oct 2013, 00:01

Expert's post

NeetiGupta wrote:

Bunuel wrote:

Bunuel wrote:

Take the given expression to the 4th power: x^4=x^3+6x^2;

Re-arrange and factor out x^2: x^2(x^2-x-6)=0;

Factorize: x^2(x-3)(x+2)=0;

So, the roots are x=0, x=3 and x=-2. But x cannot be negative as it equals to the even (4th) root of some expression (\sqrt{expression}\geq{0}), thus only two solution are valid x=0 and x=3.

The sum of all possible solutions for x is 0+3=3.

Answer: D.

Hi Bunuel, Request you to please let me know where I'm making a mistake.

Since x^3+6*x^2 >= 0 Then, x^2(x+6)>=0

i.e (x-0)^2(x-(-6))>=0 This implies thatx>=-6 Hence, x=-2 is a valid root, and sum of all roots should bex=3+(-2) = 1 Please let me know where I am going wrong.

Thanks

Plug x=-2 into x=\sqrt[4]{x^3+6x^2}. Does the equation hold true?

Hi Brunel if say x=16. x^1/2 has two values +4 & -4. So why x^1/4 cannot have +2 & -2 as as values ? Please explain

First of all it's 4th root not 2nd root. Next, \sqrt[4]{16}=2, not +2 or -2.

When the GMAT provides the square root sign for an even root, such as \sqrt{x} or \sqrt[4]{x}, then the only accepted answer is the positive root. That is, \sqrt{25}=5, NOT +5 or -5.

In contrast, the equation x^2=25 has TWO solutions, +5 and -5. Even roots have only non-negative value on the GMAT.

Re: New Algebra Set!!! [#permalink]
05 Apr 2014, 02:42

Q1. Option D. Raising both LHS and RHS to fourth power,and making RHS=0 we get x^4-x^3-6x^2=0 Taking x^2 common and factorizing, x^2(x+2)(x-3)=0 Three possible values of x are -2,3 and 0 But x can't be -ve. So sum of possible values=3

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