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New Algebra Set!!!

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WoundedTiger
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Re: New Algebra Set!!! [#permalink] New post   21 Jun 2014, 10:37
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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: \(x^4=x^3+6x^2\);

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.


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:
math-number-theory-88376.html#p666609
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Re: New Algebra Set!!! [#permalink] New post   06 Aug 2014, 11:00
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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:
https://magoosh.com/gmat/2012/foil-on-th ... expanding/
https://magoosh.com/gmat/2012/algebra-on ... to-factor/

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^2
Square of a difference: (a − b)^2 = a^2 − 2ab + b^2
See:
https://magoosh.com/gmat/2013/three-alge ... -the-gmat/

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!!! [#permalink] New post   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: \(x^4=x^3+6x^2\);

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,

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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Re: New Algebra Set!!! [#permalink] New post   28 May 2015, 04:18
Expert Reply
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: \(x^4=x^3+6x^2\);

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,

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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Re: New Algebra Set!!! [#permalink] New post   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

\(-3x^2 + 12x -2y^2 - 12y - 39=-3x^2 + 12x-12-2y^2 - 12y-18-9=-3(x^2-4x+4)-2(y^2+6y+9)-9=\)
\(=-3(x-2)^2-2(y+3)^2-9\).

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.


Don't understand why to to maximize the value of \(-3(x-2)^2-2(y+3)^2\) must be Zero ??
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Re: New Algebra Set!!! [#permalink] New post   31 Jan 2017, 10:52
Expert Reply
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

\(-3x^2 + 12x -2y^2 - 12y - 39=-3x^2 + 12x-12-2y^2 - 12y-18-9=-3(x^2-4x+4)-2(y^2+6y+9)-9=\)
\(=-3(x-2)^2-2(y+3)^2-9\).

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.


Don't understand why to to maximize the value of \(-3(x-2)^2-2(y+3)^2\) must be Zero ??


The max value of -3(x-2)^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!!! [#permalink] New post   12 Dec 2017, 08:53
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

Re-arrange and factor for x^2: \((x^2-25)(x^2-4)=0\).

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.


How did you get this directly?
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Re: New Algebra Set!!! [#permalink] New post   12 Dec 2017, 08:57
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srishti201996 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

Re-arrange and factor for x^2: \((x^2-25)(x^2-4)=0\).

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.


How did you get this directly?


Factoring Quadratics
Solving Quadratic Equations

For more:

7. Algebra



Other topics: Ultimate GMAT Quantitative Megathread

Hope it helps.
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Re: New Algebra Set!!! [#permalink] New post   06 Feb 2018, 10:21
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: \(x^2-8x-b=0\).

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,

hope you can help:

In the description it says "has equal roots, and one of the roots of the equation is ... 3".

I know that when an equation has only one solution, the determinant needs to be zero. BUT: in the description it says "and one of the roots is ... 3" --> therefore i would have never thought about setting the determinant equal to zero.

Could you please explain where the hint is to know that it's necessary to set the determinant equal to zero?

Thank you very much!

Best regards
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Re: New Algebra Set!!! [#permalink] New post   06 Feb 2018, 10:24
Expert Reply
andr3 wrote:
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: \(x^2-8x-b=0\).

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,

hope you can help:

In the description it says "has equal roots, and one of the roots of the equation is ... 3".

I know that when an equation has only one solution, the determinant needs to be zero. BUT: in the description it says "and one of the roots is ... 3" --> therefore i would have never thought about setting the determinant equal to zero.

Could you please explain where the hint is to know that it's necessary to set the determinant equal to zero?

Thank you very much!

Best regards


1. x^2 + ax - b = 0 has equal roots

2. One of the roots of the equation x^2 + ax + 15 = 0 is 3.

Those are two different equations.
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Re: New Algebra Set!!! [#permalink] New post   06 Feb 2018, 10:42
Bunuel wrote:
andr3 wrote:
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: \(x^2-8x-b=0\).

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,

hope you can help:

In the description it says "has equal roots, and one of the roots of the equation is ... 3".

I know that when an equation has only one solution, the determinant needs to be zero. BUT: in the description it says "and one of the roots is ... 3" --> therefore i would have never thought about setting the determinant equal to zero.

Could you please explain where the hint is to know that it's necessary to set the determinant equal to zero?

Thank you very much!

Best regards


1. x^2 + ax - b = 0 has equal roots

2. One of the roots of the equation x^2 + ax + 15 = 0 is 3.

Those are two different equations.


ok, than it is that language of "has equal roots", which I don't understand. could you clarify what "equal roots" means? is "equal roots" the definition of having only one solution?

thanks!
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Re: New Algebra Set!!! [#permalink] New post   06 Feb 2018, 10:44
Expert Reply
andr3 wrote:

ok, than it is that language of "has equal roots", which I don't understand. could you clarify what "equal roots" means? is "equal roots" the definition of having only one solution?

thanks!


Yes, equal roots means that there are two roots which are equal to each other, so one distinct root.
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Re: New Algebra Set!!! [#permalink] New post   22 Feb 2018, 22:47
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

\(-3x^2 + 12x -2y^2 - 12y - 39=-3x^2 + 12x-12-2y^2 - 12y-18-9=-3(x^2-4x+4)-2(y^2+6y+9)-9=\)
\(=-3(x-2)^2-2(y+3)^2-9\).

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.


can someone help me how the max value of the highlighted text is defined ?
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Re: New Algebra Set!!! [#permalink] New post   22 Feb 2018, 22:53
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SaravanaPrabu090492 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

\(-3x^2 + 12x -2y^2 - 12y - 39=-3x^2 + 12x-12-2y^2 - 12y-18-9=-3(x^2-4x+4)-2(y^2+6y+9)-9=\)
\(=-3(x-2)^2-2(y+3)^2-9\).

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.


can someone help me how the max value of the highlighted text is defined ?


\(-3(x-2)^2= (-3)*(square \ of \ a \ number)= (negative)*(non-negative) = (non-positive)\). So, this term is negative or 0.

The same for \(-2(y+3)^2\).
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Re: New Algebra Set!!! [#permalink] New post   22 Feb 2018, 23:42
Bunuel wrote:
10. 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

\(f(x) = 2x - 1\), hence \(f(n^2)=2n^2-1\).
\(g(x) = x^2\), hence \(g(n+12)=(n+12)^2=n^2+24n+144\).

Since given that \(f(n^2)=g(n+12)\), then \(2n^2-1=n^2+24n+144\). Re-arranging gives \(n^2-24n-145=0\).

Next, Viete's theorem states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):

\(x_1+x_2=\frac{-b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).

Thus according to the above \(n_1*n_2=-145\).

Answer: A.


Having a doubt with the above highlighted text. for the first, shouldn't the whole term [2x-1] be squared like the below one ? seeking solutions.
thanks in advance.
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Re: New Algebra Set!!! [#permalink] New post   23 Feb 2018, 00:15
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SaravanaPrabu090492 wrote:
Bunuel wrote:
10. 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

\(f(x) = 2x - 1\), hence \(f(n^2)=2n^2-1\).
\(g(x) = x^2\), hence \(g(n+12)=(n+12)^2=n^2+24n+144\).

Since given that \(f(n^2)=g(n+12)\), then \(2n^2-1=n^2+24n+144\). Re-arranging gives \(n^2-24n-145=0\).

Next, Viete's theorem states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):

\(x_1+x_2=\frac{-b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).

Thus according to the above \(n_1*n_2=-145\).

Answer: A.


Having a doubt with the above highlighted text. for the first, shouldn't the whole term [2x-1] be squared like the below one ? seeking solutions.
thanks in advance.


No.

\(f(x) = 2x - 1\) means that we have a function which tells us what to do to get the value of the function: take a value for which you want to find the output of the function, multiply it by 2 and subtract 1.

So, to get \(f(n^2)\) put n^2 instead of x: \(f(n^2)=2n^2-1\).

Similarly for \(g(x) = x^2\). To get \(g(n+12)\) put n + 12 instead of x: \(g(n+12)=(n+12)^2=n^2+24n+144\).
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Re: New Algebra Set!!! [#permalink] New post   19 May 2020, 00:33
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

Re-arrange and factor for x^2: \((x^2-25)(x^2-4)=0\).

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: new-algebra-set-149349-60.html#p1200948

x^4=x^3 + 6x^2
x^4 - x^3 - 6x^2 = 0 ------------> x^2 (x^2 - x - 6) = 0 -----------> x^2 (x-3)(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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Re: New Algebra Set!!! [#permalink] New post   19 May 2020, 00:36
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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

Re-arrange and factor for x^2: \((x^2-25)(x^2-4)=0\).

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: new-algebra-set-149349-60.html#p1200948

x^4=x^3 + 6x^2
x^4 - x^3 - 6x^2 = 0 ------------> x^2 (x^2 - x - 6) = 0 -----------> x^2 (x-3)(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!!! [#permalink] New post   19 May 2020, 05:52
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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 non-negative 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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Re: New Algebra Set!!! [#permalink] New post   01 Dec 2020, 04:45
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

Re-arrange and factor for x^2: \((x^2-25)(x^2-4)=0\).

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.



Hi Bunuel, Can you please explain how did you rearrange the equation into a quadratic equation? Thanks
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Re: New Algebra Set!!! [#permalink]
01 Dec 2020, 04:45
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