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Re: New Algebra Set!!!
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20 Nov 2014, 07:36
hanschris5 wrote: Bunuel wrote: imhimanshu wrote: 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 No, it does NOT hold. You should plug into the original equation as I mention in my post you quote: \(2\neq{\sqrt[4]{(2)^3+6(2)^2}}\).
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Re: New Algebra Set!!!
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20 Nov 2014, 09:45
These questions are not from GMATPrep Exam Pack 1. I know #6 is an exact copy of an official GMAT questions, but the others are not official GMAT questions.
Dabral



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20 Nov 2014, 09:47
dabral wrote: These questions are not from GMATPrep Exam Pack 1. I know #6 is an exact copy of an official GMAT questions, but the others are not official GMAT questions.
Dabral Those are GMAT Club Tests question. Changed the tag. Thanks.
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Re: New Algebra Set!!!
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20 Nov 2014, 13:57
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. Hello, Can you pls tell me my error? I took \(x^2\) common and got \(x^2(x+6)\) = \(x^4\) then i divided everything by \(x^2\). So I only get two roots. x=2 and x=3. I don't get x=0...



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Re: New Algebra Set!!!
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20 Nov 2014, 14:22
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. Again, I did something basic wrong. I left 12 on the RHS, and took mn common to have mn(mn+1)=12. mn=12 or mn=11....What is wrong with my concept?



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20 Nov 2014, 23:30
usre123 By dividing both sides by \(x^2\) you have eliminated the solution \(x=0\). For example, if we are given the condition \(x(x1)=0\), then there are two values of \(x\) that satisfies this equation, \(x=0\) and \(x=1\). However, if we divide both sides by \(x\) we end up with \(x1=0\), and left with only one solution. In general, on the GMAT never divide by a variable, unless you know that it is nonzero. Cheers, Dabral



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20 Nov 2014, 23:34
"Again, I did something basic wrong. I left 12 on the RHS, and took mn common to have mn(mn+1)=12. mn=12 or mn=11....What is wrong with my concept? "
You will find the mn=12 or mn=11 do not satisfy the above equation. We are looking for the same value of mn, this one is easy enough that many people can guess that mn=3 works, because (3)(3+1)=12, and the other solution is mn=4, which is a bit harder to spot. The better approach is to write it as a quadratic and then factor as shown in the solution.
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Re: New Algebra Set!!!
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21 Nov 2014, 19:16
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, It is understood that even root of a negative expression is not possible. But here, for x= 2, the expression under root, x^3+6x^2 = 8 + 24 = 16 which is positive. Then why x = 2 invalid ? Please explain



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Re: New Algebra Set!!!
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26 May 2015, 11:21
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 ?



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27 May 2015, 03:28
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 ? \((mn)^2=(mn)(mn)=m^2*n^2\).
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27 May 2015, 13:52
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. Hi Bunuel, I understand how you got to the three roots: 0, 3, and 2 but I missed out on the 0 root because I factored the expression like this: x^4 = x^3 + 6x^2 x^4 = x^2(x + 6) x^2 = x + 6 x^2  x  6 = 0 .... leading to only roots 3 and 2 I'm not sure why my way is incorrect since. I don't know if I broke some algebra rule. Could you please clarify my factoring method? Thanks, kindly.



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28 May 2015, 04:18
torontoclub15 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. Hi Bunuel, I understand how you got to the three roots: 0, 3, and 2 but I missed out on the 0 root because I factored the expression like this: x^4 = x^3 + 6x^2 x^4 = x^2(x + 6) x^2 = x + 6 x^2  x  6 = 0 .... leading to only roots 3 and 2 I'm not sure why my way is incorrect since. I don't know if I broke some algebra rule. Could you please clarify my factoring method? Thanks, kindly. If you divide (reduce) x^4 = x^2(x + 6) by x^2, you assume, with no ground for it, that x^2 (x) does not equal to zero thus exclude a possible solution. Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We cannot divide by zero.
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11 Sep 2015, 08:17
Bunuel wrote: Sorry, there was a typo in the stem .
5. 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
Rearrange the given equation: \(x^22x+15=m\).
Given that \(x\) is an integer from 10 and 10, inclusive (21 values) we need to find the probability that \(x^22x+15\) is greater than zero, so the probability that \(x^22x+15>0\).
Factorize: \((x+5)(3x)>0\). This equation holds true for \(5<x<3\).
Since x is an integer then it can take the following 7 values: 4, 3, 2, 1, 0, 1, and 2.
So, the probability is 7/21=1/3.
Answer: B. Hi Bunuel Could you please check below is fine? Thanks can we write ? m>0 > x^22x+15 > 0 x^2+2x15 < 0 (x+5)(x3) < 0 test some big number like 100 > (x+5)(x3) will be positive  +ve 5  ve 3 +ve range of x ==> 5<x<3



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Re: New Algebra Set!!!
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11 Sep 2015, 22:18
Hi anupamadw,
Yes, that works. When you have a quadratic inequality and there are two real roots, the number line can be thought of being divided in to three regions. One to the left of x=5, the region between 5 and 3, and the region to the right of 3. The sign of the expression will change as we go from one region to the next. This means our expression will either follow the path of +, , + or , +,  in the three regions. When you chose x=100 as a sample value, we found that the expression (x+5)(x3) is positive at x=100 and we can conclude that for x>3 it is positive, this means that between 5 and 3 it will be negative, and less 5 it will be positive.
Cheers, Dabral



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Re: New Algebra Set!!!
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28 Jan 2017, 04:23
WoundedTiger 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
Sol: Consider eqn 2 first. Since 3 is the root of the eqn we get
3^2 +3a+15=0> Solving for a we get a=8.
Putting the value if a in Eqn 1 we get x^2 8xb=0 We know for Eqn ax^2 +bx+c=0 Sum of roots is given by b/a Product of roots is c/a
Therefore from Eqn1 (after subsitution of value of a) we get Sum of roots = 8 and roots are equal ie. 4 and 4 Product of roots will be 16. Therefore Option B Hello How do I know that I have to plug in 3 for X if it says that the root of the equation is 3. I dont understand. Can someone explain? Thank you very much



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Re: New Algebra Set!!!
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28 Jan 2017, 07:52
sonnenblume92 wrote: WoundedTiger 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
Sol: Consider eqn 2 first. Since 3 is the root of the eqn we get
3^2 +3a+15=0> Solving for a we get a=8.
Putting the value if a in Eqn 1 we get x^2 8xb=0 We know for Eqn ax^2 +bx+c=0 Sum of roots is given by b/a Product of roots is c/a
Therefore from Eqn1 (after subsitution of value of a) we get Sum of roots = 8 and roots are equal ie. 4 and 4 Product of roots will be 16. Therefore Option B Hello How do I know that I have to plug in 3 for X if it says that the root of the equation is 3. I dont understand. Can someone explain? Thank you very much 3 is a root of x^2 + ax + 15 = 0 means that when you plug 3 there the equation must hold true.
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Re: New Algebra Set!!!
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29 Jan 2017, 13:20
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. Don't understand why to to maximize the value of \(3(x2)^22(y+3)^2\) must be Zero ??



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31 Jan 2017, 10:52
coolkl 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. Don't understand why to to maximize the value of \(3(x2)^22(y+3)^2\) must be Zero ?? The max value of 3(x2)^2 is 0 and max value of 2(y+3)^2 is 0 too. You see since the square of a number is always non negative, then 2*(y+3)^2 = negative*nonnegative = nonposiitve (so 2(y+3)^2 is negative or zero).
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Re: New Algebra Set!!!
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31 Jan 2017, 20:08
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 Bunnel, As per the explanation the maximum value of the expresssion is zero because the expressions in the brackets is squared (implying it will always be positive ) and the expression is multiplied by a negative number. But what if these were the expressions: \(3(x2)^2\) and \(2(y+3)^2\) How can we find the maximum values of such expressions? Thanks. Adi



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01 Feb 2017, 08:09
adityapareshshah 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 Bunnel, As per the explanation the maximum value of the expresssion is zero because the expressions in the brackets is squared (implying it will always be positive ) and the expression is multiplied by a negative number. But what if these were the expressions: \(3(x2)^2\) and \(2(y+3)^2\) How can we find the maximum values of such expressions? Thanks. Adi In this case there won't be a maximum value.
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