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

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

niks18 · Accepted Answer

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

MathRevolution · Answer

=>

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

Xin Cho · Answer

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

gvij2017 · Answer

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 · Answer

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/

niks18 · Answer

hi  Xin  Cho

pls see my post above for mathematical derivation. But for GMAT it is not required

himanshu1105 · Answer

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')

aanchals · Answer

What is the maximum value of a quadratic equation?

vaibhav1221 · Answer

Using differentiation, I think this a 30 second question. 
 \frac{dy}{dx} = 2px - 4 = 0 
 x = \frac{4}{2p}

We just need the value of p.

NotFine233 · Answer

So the question asks for the value of "x", so how does one know that the question is asking for the minimum value?

I chose E because I didn't think the question specifically asked for the minimum value, only the value of x

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

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