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The function y=px^2-4x+q in the x-y plane attains a

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New post Updated on: 09 Jan 2018, 18:53
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[GMAT math practice question]

The function \(y=px^2-4x+q\) in the \(xy\)-plane attains a minimum value. What is the value of \(x\)?

1) \(p = 2\)
2) \(q = 5\)

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Originally posted by MathRevolution on 08 Jan 2018, 01:26.
Last edited by MathRevolution on 09 Jan 2018, 18:53, edited 2 times in total.
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Re: The function y=px^2-4x+q in the x-y plane attains a  [#permalink]

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New post 20 Aug 2018, 00:34
2
Well if you have knowledge of differentiation, this question could be solved easily.

Given y = px^2−4x+qy
by differentiating :
dy/dx = 2px -4 = 0 (At lowermost point, the slope of the curve is Zero that's why dy/dx (slope) =0 )
x=4/2p => x= 2/p ( now you just need the value of 'p')
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Re: The function y=px^2-4x+q in the x-y plane attains a  [#permalink]

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New post 14 Aug 2018, 10:16
1
Xin Cho wrote:
MathRevolution wrote:
=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question.
\(y=px^2-4x+q\) has a minimum value when \(x = \frac{-(-4)}{2p} = \frac{2}{p}.\)
Thus, the question asks for the value of \(p\).

Since only condition 1) gives us information about p, only condition 1) is sufficient.
Therefore, A is the answer.
Answer: A


Can you please explain how did you derive the minimum value? Frankly, I still don't understand.


hi Xin Cho

pls see my post above for mathematical derivation. But for GMAT it is not required
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Re: The function y=px^2-4x+q in the x-y plane attains a  [#permalink]

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New post 08 Jan 2018, 01:39
MathRevolution wrote:
[GMAT math practice question]

The function \(y=px^2-4x+q\) in the \(x-y\) plane attains a minimum value. What is the value of \(x\)?

1) \(p = 2\)
2) \(q = 5\)


Minimum value of a quadratic equation is at \(\frac{-b}{2a}\)

here \(b=-4\) and \(a=p\), so minimum \(x=\frac{4}{2p}\). We need the value of \(p\) to find the minimum value.

Statement 1: \(p=2\). Sufficient

Statement 2: value of \(p\) is not mentioned. Insufficient

Option A
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Re: The function y=px^2-4x+q in the x-y plane attains a  [#permalink]

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New post 10 Jan 2018, 01:00
=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question.
\(y=px^2-4x+q\) has a minimum value when \(x = \frac{-(-4)}{2p} = \frac{2}{p}.\)
Thus, the question asks for the value of \(p\).

Since only condition 1) gives us information about p, only condition 1) is sufficient.
Therefore, A is the answer.
Answer: A
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Re: The function y=px^2-4x+q in the x-y plane attains a  [#permalink]

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New post 13 Aug 2018, 13:10
MathRevolution wrote:
=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question.
\(y=px^2-4x+q\) has a minimum value when \(x = \frac{-(-4)}{2p} = \frac{2}{p}.\)
Thus, the question asks for the value of \(p\).

Since only condition 1) gives us information about p, only condition 1) is sufficient.
Therefore, A is the answer.
Answer: A


Can you please explain how did you derive the minimum value? Frankly, I still don't understand.
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Re: The function y=px^2-4x+q in the x-y plane attains a  [#permalink]

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New post 13 Aug 2018, 18:25
I have one question here.
How did you assume that both roots are equal?

-b/2a is minimum value of quadratic equation when b^2-4ac is equal to 0 and a>0; Both roots are equal.

What do you say?

niks18 wrote:
MathRevolution wrote:
[GMAT math practice question]

The function \(y=px^2-4x+q\) in the \(x-y\) plane attains a minimum value. What is the value of \(x\)?

1) \(p = 2\)
2) \(q = 5\)


Minimum value of a quadratic equation is at \(\frac{-b}{2a}\)

here \(b=-4\) and \(a=p\), so minimum \(x=\frac{4}{2p}\). We need the value of \(p\) to find the minimum value.

Statement 1: \(p=2\). Sufficient

Statement 2: value of \(p\) is not mentioned. Insufficient

Option A
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The function y=px^2-4x+q in the x-y plane attains a  [#permalink]

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New post 14 Aug 2018, 10:13
gvij2017 wrote:
I have one question here.
How did you assume that both roots are equal?

-b/2a is minimum value of quadratic equation when b^2-4ac is equal to 0 and a>0; Both roots are equal.

What do you say?

niks18 wrote:
MathRevolution wrote:
[GMAT math practice question]

The function \(y=px^2-4x+q\) in the \(x-y\) plane attains a minimum value. What is the value of \(x\)?

1) \(p = 2\)
2) \(q = 5\)


Minimum value of a quadratic equation is at \(\frac{-b}{2a}\)

here \(b=-4\) and \(a=p\), so minimum \(x=\frac{4}{2p}\). We need the value of \(p\) to find the minimum value.

Statement 1: \(p=2\). Sufficient

Statement 2: value of \(p\) is not mentioned. Insufficient

Option A


hi gvij2017

a quadratic equation is of the form \(y=ax^2+bx+c\), it is a parabola and the value of \(a\) decides whether the parabola opens upwards or downwards

the maximum/minimum value of the equation will be at \(\frac{dy}{dx}=0\), so on differentiating the quadratic equation you will get

\(\frac{dy}{dx}=2ax+b=0 =>\frac{dy}{dx}=\frac{-b}{2a}\)

Hence the equation will have minimum/maximum point at \(\frac{-b}{2a}\)

I have used calculus here but the relation can be derived using simple algebra also. check out the below link for detailed explanations -

https://brilliant.org/wiki/maximum-valu ... -equation/
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The function y=px^2-4x+q in the x-y plane attains a   [#permalink] 14 Aug 2018, 10:13
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