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

29

This post received KUDOS

Expert's post

46

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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]
13 Apr 2013, 08:01

Expert's post

Hi Anshuman,

For this you need to go thru some strategy books of GMAT Quant. In My Opinion, two important sources are, 1) Manhattan GMAT Quant strategy guides 4th or 5th Ed 2) GMAT Club Math Book - This is free and you can download it from this website free of cost (Check BB's profile) Furthermore you can check my signature. There i have attached my articles prepared on some peculiar topics. You may find them useful.

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 that\(x>=-6\) Hence, x=-2 is a valid root, and sum of all roots should be\(x=3+(-2) = 1\) 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 that\(x>=-6\) Hence, x=-2 is a valid root, and sum of all roots should be\(x=3+(-2) = 1\) 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 that\(x>=-6\) Hence, x=-2 is a valid root, and sum of all roots should be\(x=3+(-2) = 1\) 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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