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New Algebra Set!!!

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Re: New Algebra Set!!!  [#permalink]

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New post 13 Mar 2017, 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.


Hi Bunuel, how come I cannot/should not solve this problem by treating them as if they are two equations and multiple unknowns? As in, x^2 + ax + 15 = 0 ; x^2 +ax -b = 0. Subtract the two equations and you get 15+b=0 so b= -15. Thanks!
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Re: New Algebra Set!!!  [#permalink]

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New post 13 Mar 2017, 08:36
flexman 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, how come I cannot/should not solve this problem by treating them as if they are two equations and multiple unknowns? As in, x^2 + ax + 15 = 0 ; x^2 +ax -b = 0. Subtract the two equations and you get 15+b=0 so b= -15. Thanks!


Because those are not simultaneous equations, those two are not given to be a system of equations.
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Re: New Algebra Set!!!  [#permalink]

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New post 18 Mar 2017, 10:38
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.


Hello Bunnel,

I did the same way as above except when
-x^2-2x+15 > 0,
so took -ve sign common as
x^2+2x-15 < 0
(x+5)(x-3) < 0
x+5 < 0 or x -3 < 0
x < -5 or x < 3. So for me possible values are -6,-7-8,-9,-10. Means 5/21.
Any idea where am i wrong ?
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Re: New Algebra Set!!!  [#permalink]

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New post 19 Mar 2017, 03:16
coolkl 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.


Hello Bunnel,

I did the same way as above except when
-x^2-2x+15 > 0,
so took -ve sign common as
x^2+2x-15 < 0
(x+5)(x-3) < 0
x+5 < 0 or x -3 < 0
x < -5 or x < 3. So for me possible values are -6,-7-8,-9,-10. Means 5/21.
Any idea where am i wrong ?


When you say that -6,-7-8,-9,-10 satisfy the inequality, did you test any of these values? Ho did you solve (x+5)(x-3) < 0 to get x < -5 or x < 3? What does x < -5 OR x < 3 even mean? It does not make sense... (x+5)(x-3) < 0 holds true for -5 < x < 3.

Check the links below for more:
Inequality tips

Inequalities Made Easy!

Solving Quadratic Inequalities - Graphic Approach

Wavy Line Method Application - Complex Algebraic Inequalities

DS Inequalities Problems
PS Inequalities Problems
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Re: New Algebra Set!!!  [#permalink]

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New post 14 Jun 2017, 20:09
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.



You have been clearing doubts , one more need your help. WHy X cannot take negative value?
What I understand that If X=-3 then (-3)^4=(3)^4 .
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Re: New Algebra Set!!!  [#permalink]

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New post 14 Jun 2017, 21:19
gmat4varun 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.



You have been clearing doubts , one more need your help. WHy X cannot take negative value?
What I understand that If X=-3 then (-3)^4=(3)^4 .


Please read the whole thread.

\(x\) 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]

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New post 12 Dec 2017, 07:53
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]

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New post 12 Dec 2017, 07:57
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]

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New post 06 Feb 2018, 09:21
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]

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New post 06 Feb 2018, 09:24
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]

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New post 06 Feb 2018, 09:42
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]

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New post 06 Feb 2018, 09:44
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]

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New post 22 Feb 2018, 21:47
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]

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New post 22 Feb 2018, 21:53
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]

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New post 22 Feb 2018, 22:42
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]

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New post 22 Feb 2018, 23:15
1
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]

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New post 05 Apr 2018, 08:40
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

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

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.


\(-3x^2 + 12x -2y^2 - 12y - 39\) -->\(x(12-3x) - y(2y+12)-39\)

now x(12-3x) will be highest when x is 1.

-y(2y+12) will be higest when y = -1.
so the answer becomes 9+10-39 = -19..
where am I going wrong?
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Re: New Algebra Set!!!  [#permalink]

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New post 05 Apr 2018, 08:51
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.



Quote:
Factorize: \((x+5)(3-x)>0\). This equation holds true for \(-5<x<3\)


unable to understand this line. Can someone please explain?
many thanks
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Re: New Algebra Set!!!  [#permalink]

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New post 05 Apr 2018, 12:27
Nixondutta 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.



Quote:
Factorize: \((x+5)(3-x)>0\). This equation holds true for \(-5<x<3\)


unable to understand this line. Can someone please explain?
many thanks


Check this: Solving Quadratic Inequalities: Graphic Approach

Fro more: check

Inequalities



For more check Ultimate GMAT Quantitative Megathread



Hope it helps.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: New Algebra Set!!! &nbs [#permalink] 05 Apr 2018, 12:27

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