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New Algebra Set!!!
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Updated on: 18 Feb 2019, 04:31
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 Solution: https://gmatclub.com/forum/newalgebra ... l#p12009482. 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: https://gmatclub.com/forum/newalgebra ... l#p12009503. 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: https://gmatclub.com/forum/newalgebra ... l#p12009564. What is the maximum value of 3x^2 + 12x 2y^2  12y  39 ?A. 39 B. 9 C. 0 D. 9 E. 39 Solution: https://gmatclub.com/forum/newalgebra ... l#p12009625. If x^2 + 2x 15 = m, where x is an integer from 10 and 10, inclusive, what is the probability that m is greater than zero?A. 2/7 B. 1/3 C. 7/20 D. 2/5 E. 3/7 Solution: https://gmatclub.com/forum/newalgebra ... l#p12009706. If mn does not equal to zero, and m^2n^2 + mn = 12, then m could be:I. 4/n II. 2/n III. 3/n A. I only B. II only C. III only D. I and II only E. I and III only Solution: https://gmatclub.com/forum/newalgebra ... l#p12009737. 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: https://gmatclub.com/forum/newalgebra ... l#p12009758. If m is a negative integer and m^3 + 380 = 381m , then what is the value of m?A. 21 B. 20 C. 19 D. 1 E. None of the above Solution: https://gmatclub.com/forum/newalgebra ... l#p12009809. If \(x=(\sqrt{5}\sqrt{7})^2\), then the best approximation of x is:A. 0 B. 1 C. 2 D. 3 E. 4 Solution: https://gmatclub.com/forum/newalgebra ... l#p120098210. If f(x) = 2x  1 and g(x) = x^2, then what is the product of all values of n for which f(n^2)=g(n+12) ?A. 145 B. 24 C. 24 D. 145 E. None of the above Solution: https://gmatclub.com/forum/newalgebra ... l#p1200987Kudos points for each correct solution!!!
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Originally posted by Bunuel on 18 Mar 2013, 06:56.
Last edited by Bunuel on 18 Feb 2019, 04:31, edited 1 time in total.
Updated.



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Re: New Algebra Set!!!
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25 Nov 2018, 01:21
shiv17 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. Can anybody help me with this solution. Why the maximum value of \(3(x2)^2\) and \(2(y+3)^2\) is zero? Please read the whole thread before posting: https://gmatclub.com/forum/newalgebra ... l#p2020076
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Re: New Algebra Set!!!
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25 Nov 2018, 12:37
putting the value of x in eq  3^2+3a+15=0 => a= 8 Now putting a= 8 in x^2 + ax  b = 0 x^2−8x−b=0.
as roots are equal => b^24ac = 0 => b = 16 (B)



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Re: New Algebra Set!!!
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03 Jun 2019, 14:42
Bunuel wrote: LaxAvenger wrote: Bunuel wrote: 6. If mn does not equal to zero, and m^2n^2 + mn = 12, then m could be:
I. 4/n II. 2/n III. 3/n
A. I only B. II only C. III only D. I and II only E. I and III only
Rearrange: \((mn)^2 + mn  12=0\).
Factorize for mn: \((mn+4)(mn3)=0\). Thus \(mn=4\) or \(mn=3\).
So, we have that \(m=\frac{4}{n}\) or \(m=\frac{3}{n}\).
Answer: E. Hello, how come m^2n^2 will become (mn)^2 ? \((mn)^2=(mn)(mn)=m^2*n^2\). Hi  I think this explains how mn^2 becomes m^2*n^2, but how does (m)^(2n^2) become (mn)^2? I put the parenthesis to show how I read the question, so perhaps my error and the error of the original questioner is in how it is written and not how it is solved. Thanks!



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03 Jun 2019, 20:35
ghostman23 wrote: Hi  I think this explains how mn^2 becomes m^2*n^2, but how does (m)^(2n^2) become (mn)^2? I put the parenthesis to show how I read the question, so perhaps my error and the error of the original questioner is in how it is written and not how it is solved. Thanks! m^2n^2 can only mean \(m^2n^2\), nothing else. If it were m^(2n^2), it would be written that way.
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New Algebra Set!!!
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08 Jun 2019, 09:04
Bunuel wrote: 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 Solution: https://gmatclub.com/forum/newalgebra ... l#p12009482. 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: https://gmatclub.com/forum/newalgebra ... l#p12009503. 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: https://gmatclub.com/forum/newalgebra ... l#p12009564. What is the maximum value of 3x^2 + 12x 2y^2  12y  39 ?A. 39 B. 9 C. 0 D. 9 E. 39 Solution: https://gmatclub.com/forum/newalgebra ... l#p12009625. If x^2 + 2x 15 = m, where x is an integer from 10 and 10, inclusive, what is the probability that m is greater than zero?A. 2/7 B. 1/3 C. 7/20 D. 2/5 E. 3/7 Solution: https://gmatclub.com/forum/newalgebra ... l#p12009706. If mn does not equal to zero, and m^2n^2 + mn = 12, then m could be:I. 4/n II. 2/n III. 3/n A. I only B. II only C. III only D. I and II only E. I and III only Solution: https://gmatclub.com/forum/newalgebra ... l#p12009737. 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: https://gmatclub.com/forum/newalgebra ... l#p12009758. If m is a negative integer and m^3 + 380 = 381m , then what is the value of m?A. 21 B. 20 C. 19 D. 1 E. None of the above Solution: https://gmatclub.com/forum/newalgebra ... l#p12009809. If \(x=(\sqrt{5}\sqrt{7})^2\), then the best approximation of x is:A. 0 B. 1 C. 2 D. 3 E. 4 Solution: https://gmatclub.com/forum/newalgebra ... l#p120098210. If f(x) = 2x  1 and g(x) = x^2, then what is the product of all values of n for which f(n^2)=g(n+12) ?A. 145 B. 24 C. 24 D. 145 E. None of the above Solution: https://gmatclub.com/forum/newalgebra ... l#p1200987Kudos points for each correct solution!!! 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
A quick way to solve this question, without having to find the value of a by substituting 3 in to equation 2 is to simply see that the second equation's constant value is 15. The roots of the second equation will have to equal 15 (think about what you do when factoring quadratic equations e.g. to find factors of ax^2 + bx + c you will look for two factors of AC that in some way add up to b). So the other factor of the second equation is 5, because 3x5=15. So we know that a will be equal to 5 + 3 = 8. Once you know that you can rewrite the first equation as x^2 + 8x  b. Now remember that equation 1 has equal roots which means it follows the form of (a+b)^2 which expands to a^2 + 2ab + b^2. Therefore a=2*1*4 (1 is the coefficient of x^2, and if a = 2 * first expression * second expression then a = 2*1* 4). Hence b = 4^2= 16 (think about the (a+b)^2 form again). The negative sign is needed so the expression follows this form.Answer is thus B 16



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Re: New Algebra Set!!!
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02 Nov 2019, 04:38
Bunuel wrote: 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
Since one of the roots of the equation \(x^2 + ax + 15 = 0\) is 3, then substituting we'll get: \(3^2+3a+15=0\). Solving for \(a\) gives \(a=8\).
Substitute \(a=8\) in the first equation: \(x^28xb=0\).
Now, we know that it has equal roots thus its discriminant must equal to zero: \(d=(8)^2+4b=0\). Solving for \(b\) gives \(b=16\).
Answer: B. Hi Bunuel, Why can I not substract x^2 + ax + 15 = 0 from x^2 + ax  b = 0 and get b= 15? Thanks! Lionila



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Re: New Algebra Set!!!
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18 May 2020, 23:33
Bunuel wrote: 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
Rearrange and factor for x^2: \((x^225)(x^24)=0\).
So, we have that \(x=5\), \(x=5\), \(x=2\), or \(x=2\).
\(50=5*(5)*2\); \(50=5*(5)*(2)\).
Only 25 is NOT a product of three possible values of x
Answer: B. In this question  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 Solution: newalgebraset14934960.html#p1200948 x^4=x^3 + 6x^2 x^4  x^3  6x^2 = 0 > x^2 (x^2  x  6) = 0 > x^2 (x3)(x+2) = 0 x=0 or 3 or 2. With x = 2 original equation does not hold true so possible values for X are 3 and 0 Hence their sum is 3 Choice D. You didn't take the negative values  why are we considering negative values here for 4th roots?



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18 May 2020, 23:36
TarPhi wrote: Bunuel wrote: 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
Rearrange and factor for x^2: \((x^225)(x^24)=0\).
So, we have that \(x=5\), \(x=5\), \(x=2\), or \(x=2\).
\(50=5*(5)*2\); \(50=5*(5)*(2)\).
Only 25 is NOT a product of three possible values of x
Answer: B. In this question  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 Solution: newalgebraset14934960.html#p1200948 x^4=x^3 + 6x^2 x^4  x^3  6x^2 = 0 > x^2 (x^2  x  6) = 0 > x^2 (x3)(x+2) = 0 x=0 or 3 or 2. With x = 2 original equation does not hold true so possible values for X are 3 and 0 Hence their sum is 3 Choice D. You didn't take the negative values  why are we considering negative values here for 4th roots? This is explained several times on previous pages. \(x\) cannot be negative as it equals to the even (4th) root of some expression (\(\sqrt{expression}\geq{0}\)). 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 a positive value on the GMAT.In contrast, the equation \(x^2=25\) has TWO solutions, +5 and 5.
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Re: New Algebra Set!!!
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18 May 2020, 23:47
Bunuel wrote: TarPhi wrote: Bunuel wrote: 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
Rearrange and factor for x^2: \((x^225)(x^24)=0\).
So, we have that \(x=5\), \(x=5\), \(x=2\), or \(x=2\).
\(50=5*(5)*2\); \(50=5*(5)*(2)\).
Only 25 is NOT a product of three possible values of x
Answer: B. In this question  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 Solution: newalgebraset14934960.html#p1200948 x^4=x^3 + 6x^2 x^4  x^3  6x^2 = 0 > x^2 (x^2  x  6) = 0 > x^2 (x3)(x+2) = 0 x=0 or 3 or 2. With x = 2 original equation does not hold true so possible values for X are 3 and 0 Hence their sum is 3 Choice D. You didn't take the negative values  why are we considering negative values here for 4th roots? This is explained several times on previous pages. \(x\) cannot be negative as it equals to the even (4th) root of some expression (\(\sqrt{expression}\geq{0}\)). 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 a positive value on the GMAT.In contrast, the equation \(x^2=25\) has TWO solutions, +5 and 5. Thanks Bunuel. Apologies for that! I kept getting confused! Even roots (sq root and 4th root) cannot have a ve value but x^2=16 and x^4=16 can have (+2 or2) as the solutions. I got it Thanks again!



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Re: New Algebra Set!!!
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19 May 2020, 04:52
Bunuel wrote: 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 a positive value on the GMAT.
I know this is not what Bunuel meant to suggest, but there is no math "on the GMAT" that is different from math anywhere else. √25 equals 5 everywhere, not only on the GMAT, because in math the radical symbol "√" is defined to mean "the nonnegative square root". I've seen people ask questions like "Is zero an even number on the GMAT?", which makes me think that some people are under the impression that there are special math conventions you need to learn for the GMAT that might be different from those you've learned in real math. That's not true: GMAT math is the same thing as real math. Zero is even everywhere, not just on the GMAT. √25 = 5 everywhere, not only on the GMAT. There's no reason to ever include the qualification "on the GMAT" when explaining what's true in math.
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