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: \(p=2\).
SufficientStatement 2: value of \(p\) is not mentioned.
InsufficientOption
Ahi
gvij2017a 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/