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New Algebra Set!!!

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Bunuel
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Re: New Algebra Set!!! [#permalink] New post   01 Dec 2020, 04:51
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Nishant1795 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.



Hi Bunuel, Can you please explain how did you rearrange the equation into a quadratic equation? Thanks


x^4 = 29x^2 - 100;

x^4 - 29x^2 + 100 = 0;

Say x^2 = a, so we have a^2 - 29a + 100 = 0.

Now, you can solve for a or factor.

Factoring Quadratics: https://www.purplemath.com/modules/factquad.htm
Solving Quadratic Equations: https://www.purplemath.com/modules/solvquad.htm

Hope it helps.
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abhinavsodha800
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Re: New Algebra Set!!! [#permalink] New post   06 Dec 2020, 10:49
Hi Bunuel,
In first question why x cannot be negative . i know the square root is always positive . but when i put the value of x =-2 it staisfy both the side of equation which is creating a confusion for me.
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Bunuel
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Re: New Algebra Set!!! [#permalink] New post   06 Dec 2020, 10:55
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abhinavsodha800 wrote:
Hi Bunuel,
In first question why x cannot be negative . i know the square root is always positive . but when i put the value of x =-2 it staisfy both the side of equation which is creating a confusion for me.


If x = -2, then \(\sqrt[4]{x^3+6x^2}=\sqrt[4]{-8+24}=\sqrt[4]{16}=2\).

As you can see \(x≠\sqrt[4]{x^3+6x^2}\).
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Re: New Algebra Set!!! [#permalink] New post   26 Aug 2022, 00:55
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.


Hi Bunuel, could you please explain this?
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Re: New Algebra Set!!! [#permalink] New post   26 Aug 2022, 00:59
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ishitaagarwal59 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.


Hi Bunuel, could you please explain this?


Sure. The square of a number is positive or 0.

So, \(-3*(x-2)^2=(negative)*(positive \ or \ 0)=(negative \ or \ 0)\), thus the max value of \(-3(x-2)^2\) is 0. The same with \(-2(y+3)^2\).

Hope it's clear.
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Re: New Algebra Set!!! [#permalink] New post   14 Mar 2023, 16:16
Bunuel are these questions available anywhere as question posts such that they can be tracked in an error log?
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Re: New Algebra Set!!! [#permalink] New post   15 Mar 2023, 00:02
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lecremeglace wrote:
Bunuel are these questions available anywhere as question posts such that they can be tracked in an error log?


Yes, they are available in GMAT Club Tests Forum but you'd need a subscription to it.
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Re: New Algebra Set!!! [#permalink] New post   05 Apr 2023, 17:33
FInd in the photo below the answer to question 2:
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Re: New Algebra Set!!! [#permalink] New post   05 Apr 2023, 19:59
Bunuel wrote:
8. 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

Given \(m^3 + 380 = 380m+m\).

Re-arrange: \(m^3-m= 380m-380\).

\(m(m+1)(m-1)=380(m-1)\). Since m is a negative integer, then \(m-1\neq{0}\) and we can safely reduce by \(m-1\) to get \(m(m+1)=380\).

So, we have that 380 is the product of two consecutive negative integers: \(380=-20*(-19)\), hence \(m=-20\).

Answer: B.



Bunuel can you please provide more questions like this one for more practice?

Please find in the photo below my answer to question 8:
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Re: New Algebra Set!!! [#permalink] New post   06 Apr 2023, 08:17
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.


Also after we know a=-8, and that roots of the first equation are equal,

sum of the roots=r+r=8/1
producto of the root= r*r=-b/1

2r=8 so r=4

r^2=-b
16=-b
-16=b
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Re: New Algebra Set!!! [#permalink] New post   28 May 2023, 17:31
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

Re-arrange the given equation: \(-x^2-2x+15=m\).

Given that \(x\) is an integer from -10 and 10, inclusive (21 values) we need to find the probability that \(-x^2-2x+15\) is greater than zero, so the probability that \(-x^2-2x+15>0\).

Factorize: \((x+5)(3-x)>0\). 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.


no...you have to include 3 and -5. both gives the value of 0. 9/21

here it is asked value greater than zero.
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Re: New Algebra Set!!! [#permalink] New post   29 May 2023, 00:12
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pudu wrote:
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

Re-arrange the given equation: \(-x^2-2x+15=m\).

Given that \(x\) is an integer from -10 and 10, inclusive (21 values) we need to find the probability that \(-x^2-2x+15\) is greater than zero, so the probability that \(-x^2-2x+15>0\).

Factorize: \((x+5)(3-x)>0\). 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.


no...you have to include 3 and -5. both gives the value of 0. 9/21

here it is asked value greater than zero.


You are wrong.

If x = 3, then x^2 + 2x -15 = 0, not greater than 0.
If x = -5, then x^2 + 2x -15 = 0, not greater than 0.
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Re: If x^2 +2x15=m , where x is an integer from -10 and 10, i [#permalink] New post   24 Sep 2023, 09:21
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Re: If x^2 +2x15=m , where x is an integer from -10 and 10, i [#permalink]
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