If y = x^2 + ax + b, y is minimum when x is:

a) a/b
b) -a/b
c) -a/2
d) -b/2
e) b/a

I tried it by substituting the value of x everywhere For once that is making the problem lengthy, secondly I got stuck . Can anybody help please?

Hi shreyashid,

This question can be solved by TESTing VALUES.

We're given the equation Y = X^2 + AX + B.

IF.. we use a simple Classic Quadratic....
A = 2
B = 1
Y = X^2 + 2X + 1

We can then go about finding the answer that yields the MINIMUM result when X = ...

Answer A: A/B = 2/1 = 2 --> 4+4+1 = +9
Answer B: -A/B = -2/1 = -2 --> 4-4+1 = +1
Answer C: -A/2 = -2/2 = -1 --> 1-2+1 = 0
Answer D: -B/2 = -1/2 -->(1/4)-1+1 = +1/4
Answer E: B/A = 1/2 --> (1/4)+1+1 = +2 1/4

From these results, we can see the minimum result:

Final Answer:

GMAT assassins aren't born, they're made,
Rich
_________________

Given that:
y=x^2+ax+b.
differentiate this equation and equate to zero.
d(y)/d(x)=2x+a.....1
equate it to zero.
2x+a=0.
x=-a/2.

The minimum value of this equation is found by differentiating it once and putting = 0
Hence 2Ax + B = 0
how did you reach to 2Ax+B=0.Pls explain in detail.srt for the troube

Here's my solution!

\(y = x^2 + ax + b\)

Considering that \(b^2 - 4ac = 0\)

\(b^2 =4ac\)

Where \(a = x^2\), \(b = a\), \(c = b\) which is constant (so can be just 1).

\(a^2 =4x^2(1)\)

\(\frac{a^2}{4} = x^2\)

\(\frac{a}{2} = x\)

OR

\(\frac{-a}{2} = x\)

After we square root we have +ve and -ve values.

C