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23 May 2013, 05:13
3. If a and b are positive numbers, such that a^2 + b^2 = m and a^2  b^2 = n, then ab in terms of m and n equals to:
A. \(\frac{\sqrt{mn}}{2}\) B. \(\frac{\sqrt{mn}}{2}\) C. \(\frac{\sqrt{m^2n^2}}{2}\) D. \(\frac{\sqrt{n^2m^2}}{2}\) E. \(\frac{\sqrt{m^2+n^2}}{2}\)
Solution:
a^2  b^2 = n => (a+b)(ab)=n=> (a+b)^2*(ab)^2=n^2 => (a^2 + b^2+2ab)*(a^2 + b^22ab)=n^2 => (m+2ab)*(m2ab)=n^2 => m^22ab^2=n^2 =>2ab^2=m^2n^2 => 2ab=\sqrt{\(m^2n^2\)} =>ab=\sqrt{\(m^2n^2\)}/2
Ans: C



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23 May 2013, 05:27
7. If x^4 = 29x^2  100, then which of the following is NOT a product of three possible values of x? I. 50 II. 25 III. 50 A. I only B. II only C. III only D. I and II only E. I and III only Solution:By using below substitution values we could reduce the powers of x. Here values (1,0,29,0,100)  taken from equation  x^4 = 29x^2  100=> (1)x^4+(0)x^329x^2+(0)x+100=0 from above explanation we get equation x^225=0 => give x=+5,5 from above calculation we get values of x =+2,2. multiplying any of the three roots we have possibility of getting product values, 50,+50. Ans:E



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04 Jun 2013, 23:45
In question no.6) how is m^2n^2=(mn)^2 ? Someone help me.



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05 Jun 2013, 00:40
pradeepkamp wrote: In question no.6) how is m^2n^2=(mn)^2 ? Someone help me. \((mn)^2 = (mn)*(mn)=m^2n^2\). Hope it's clear.
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23 Jun 2013, 00:35
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"?



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23 Jun 2013, 01:37
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 nonnegative value on the GMAT.In contrast, the equation \(x^2=25\) has TWO solutions, \(\sqrt{25}=+5\) and \(\sqrt{25}=5\). Hope it helps.
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Re: New Algebra Set!!!
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21 Jul 2013, 22:05
Bunuel wrote: 4. What is the maximum value of 3x^2 + 12x 2y^2  12y  39 ?
A. 39 B. 9 C. 0 D. 9 E. 39
\(3x^2 + 12x 2y^2  12y  39=3x^2 + 12x122y^2  12y189=3(x^24x+4)2(y^2+6y+9)9=\) \(=3(x2)^22(y+3)^29\).
So, we need to maximize the value of \(3(x2)^22(y+3)^29\).
Since, the maximum value of \(3(x2)^2\) and \(2(y+3)^2\) is zero, then the maximum value of the whole expression is \(0+09=9\).
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 !!



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21 Jul 2013, 22:09
targetgmatchotu wrote: Bunuel wrote: 4. What is the maximum value of 3x^2 + 12x 2y^2  12y  39 ?
A. 39 B. 9 C. 0 D. 9 E. 39
\(3x^2 + 12x 2y^2  12y  39=3x^2 + 12x122y^2  12y189=3(x^24x+4)2(y^2+6y+9)9=\) \(=3(x2)^22(y+3)^29\).
So, we need to maximize the value of \(3(x2)^22(y+3)^29\).
Since, the maximum value of \(3(x2)^2\) and \(2(y+3)^2\) is zero, then the maximum value of the whole expression is \(0+09=9\).
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 !! I completed the squares for 3x^2 + 12x  ... and for 2y^2  12y... So, I asked myself what do I need there in order to have (a+b)^2. Hope it's clear.
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23 Oct 2013, 23:54
Bunuel wrote: imhimanshu wrote: Bunuel wrote: [ Take the given expression to the 4th power: \(x^4=x^3+6x^2\);
Rearrange and factor out x^2: \(x^2(x^2x6)=0\);
Factorize: \(x^2(x3)(x+2)=0\);
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 \((x0)^2(x(6))>=0\) 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



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24 Oct 2013, 01:01
NeetiGupta wrote: Bunuel wrote: Bunuel wrote: Take the given expression to the 4th power: \(x^4=x^3+6x^2\);
Rearrange and factor out x^2: \(x^2(x^2x6)=0\);
Factorize: \(x^2(x3)(x+2)=0\);
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 \((x0)^2(x(6))>=0\) 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 nonnegative value on the GMAT.Hope it helps.
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Re: New Algebra Set!!!
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21 Jun 2014, 01:35
Bunuel wrote: SOLUTIONs:
1. If \(x=\sqrt[4]{x^3+6x^2}\), then the sum of all possible solutions for x is:
A. 2 B. 0 C. 1 D. 3 E. 5
Take the given expression to the 4th power: \(x^4=x^3+6x^2\);
Rearrange and factor out x^2: \(x^2(x^2x6)=0\);
Factorize: \(x^2(x3)(x+2)=0\);
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. Bunuel, how do we know that both sides of the equation are positive? if x is a negativve then we cannot raise even power.
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21 Jun 2014, 03:56
Ergenekon wrote: Bunuel wrote: SOLUTIONs:
1. If \(x=\sqrt[4]{x^3+6x^2}\), then the sum of all possible solutions for x is:
A. 2 B. 0 C. 1 D. 3 E. 5
Take the given expression to the 4th power: \(x^4=x^3+6x^2\);
Rearrange and factor out x^2: \(x^2(x^2x6)=0\);
Factorize: \(x^2(x3)(x+2)=0\);
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. Bunuel, how do we know that both sides of the equation are positive? if x is a negativve then we cannot raise even power. x cannot be negative cannot be negative as it equals to the even (4th) root of some expression.
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Re: New Algebra Set!!!
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21 Jun 2014, 10:37
Ergenekon wrote: Bunuel wrote: SOLUTIONs:
1. If \(x=\sqrt[4]{x^3+6x^2}\), then the sum of all possible solutions for x is:
A. 2 B. 0 C. 1 D. 3 E. 5
Take the given expression to the 4th power: \(x^4=x^3+6x^2\);
Rearrange and factor out x^2: \(x^2(x^2x6)=0\);
Factorize: \(x^2(x3)(x+2)=0\);
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. Bunuel, how do we know that both sides of the equation are positive? if x is a negativve then we cannot raise even power. From GMATCLUB math book: General rules: • \(\sqrt{x}\sqrt{y}=\sqrt{xy}\) and \(\frac{\sqrt{x}}{\sqrt{y}}=\sqrt{\frac{x}{y}}\). • \((\sqrt{x})^n=\sqrt{x^n}\) • \(x^{\frac{1}{n}}=\sqrt[n]{x}\) • \(x^{\frac{n}{m}}=\sqrt[m]{x^n}\) • \({\sqrt{a}}+{\sqrt{b}}\neq{\sqrt{a+b}}\) • \(\sqrt{x^2}=x\), when \(x\leq{0}\), then \(\sqrt{x^2}=x\) and when \(x\geq{0}\), then \(\sqrt{x^2}=x\) • 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 a positive value on the GMAT.• Odd roots will have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{64} =4\). More on this can be found here: mathnumbertheory88376.html#p666609
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10 Jul 2014, 02:56
Bunuel wrote: SOLUTIONs:
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 \(x=3\).
Can sumone plz explain the part in the quote? I dont quite get it I read WoundedTiger's post (one above) as well i still do not get it a lil help in layman terms plz If the entire term under the 4th sq root , say is 16 then my x can be a 2 ryt?
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10 Jul 2014, 03:07
janxavier wrote: Bunuel wrote: SOLUTIONs:
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 \(x=3\).
Can sumone plz explain the part in the quote? I dont quite get it I read WoundedTiger's post (one above) as well i still do not get it a lil help in layman terms plz If the entire term under the 4th sq root , say is 16 then my x can be a 2 ryt? \(\sqrt[4]{16}=2\), 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 nonnegative value on the GMAT.Hope it helps.
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03 Aug 2014, 08:11
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) −116 e) −164



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05 Aug 2014, 11:40
Bunuel wrote: 4. What is the maximum value of 3x^2 + 12x 2y^2  12y  39 ?
A. 39 B. 9 C. 0 D. 9 E. 39
[m]3x^2 + 12x 2y^2  12y  39=3x^2 + 12x122y^2  12y189
How do you know to break the 39 up into 12, 18, and 9?



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06 Aug 2014, 11:00
vik09 wrote: 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) −116 e) −164 Dear vik09, This is a tricky question, and I am happy to help. First of all, you may find these blogs helpful: http://magoosh.com/gmat/2012/foilonth ... expanding/http://magoosh.com/gmat/2012/algebraon ... tofactor/Let's start with x^2 + ax + 15 = 0. If one root is 3, then one factor must be (x − 3). Let's say the other root is [b]k[/b], and the other factor is (x − k). If we multiply (x  3)(x  k), the constant term will be (3k), which has to equal 15. Therefore, k = 5, and (x − 3)(x − 5) = x^2 − 8x + 15 Therefore, a = 8. Now, we have x^2 + 8x − b = 0, which has equal roots. If a quadratic has equal roots, it must be the square of either a sum or a difference: Square of a sum: (a + b)^2 = a^2 + 2ab + b^2Square of a difference: (a − b)^2 = a^2 − 2ab + b^2See: http://magoosh.com/gmat/2013/threealge ... thegmat/Look at the square of a sum formula. We replace a = x. In order for the middle term to equal 8x, we must have b = 4, and the full equation would be (x + 4)^2 = x^2 + 8x + 16 Setting the final terms equal, we have − b = 16 b = −16 Answer = (B)Does all this make sense? Mike
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Re: New Algebra Set!!!
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28 Sep 2014, 13:53
Bunuel wrote: 4. What is the maximum value of 3x^2 + 12x 2y^2  12y  39 ?
A. 39 B. 9 C. 0 D. 9 E. 39
\(3x^2 + 12x 2y^2  12y  39=3x^2 + 12x122y^2  12y189=3(x^24x+4)2(y^2+6y+9)9=\) \(=3(x2)^22(y+3)^29\).
So, we need to maximize the value of \(3(x2)^22(y+3)^29\).
Since, the maximum value of \(3(x2)^2\) and \(2(y+3)^2\) is zero, then the maximum value of the whole expression is \(0+09=9\).
Answer: B. Is there another solution in which we don't need to manipulate the number so much? How do you know when there is a shortcut like used here and when there isn't?



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Re: New Algebra Set!!!
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19 Nov 2014, 13:57
Bunuel wrote: imhimanshu wrote: Bunuel wrote: [ Take the given expression to the 4th power: \(x^4=x^3+6x^2\);
Rearrange and factor out x^2: \(x^2(x^2x6)=0\);
Factorize: \(x^2(x3)(x+2)=0\);
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 \((x0)^2(x(6))>=0\) 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? well u can plug in 2 and still it holds. x^4 = (2)^4 = 16 on the other side of equal to ie x^3+6x^2 we can write (2)^3+6(2)^2 equals 8+24 = 16 .....so holds ....please correct me . Also BTW in solving quadratic equation such as x^2x+6=0 we do consider both the roots ve and +ve , until its stated not to ... thanks




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