If y = x2 + ax + b, y is minimum when x is:

Question

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?

chetan2u · Accepted Answer

Hi,
the easiest way is algebrically...
if your quad eq is a*x^2 + b*x +c, its min or max value will occur at -b/2a...
since the eq forms a parabola/ curve due to value a*x^2, the max or min value will depend on a..
here the eq is x^2 + ax + c=y..
so b in  -b/2a... = a in  x^2 + ax + c  and a=1..
so -b/2a will become -a/2 for this equation ..
ans C..

EMPOWERgmatRichC · Answer

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:  C

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

TeamGMATIFY · Answer

Theory:
Assume the equation to be A x^2  + Bx + C = 0
The minimum value of this equation is found by differentiating it once and putting = 0
Hence 2Ax + B = 0
Therefore the minimum value of this equation will be at x =  \frac{-B}{{2A}}

Coming back to the problem, if we compare x^2 + ax + b with A x^2  + Bx + C = 0
we have A = 1, B = a and C = b

Therefore the minimum value will be at x = -a/2*1 = -a/2

Option C

senapatitanmay · Answer

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.

shreyashid · Answer

thanks everyone. the Parabola logic makes it look so simple

sanaexam · Answer

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

ganand · Answer

Hi,

We can use the method of completing the square to solve this problem.
 \begin{align*}
y &= x^{2} + ax +b\
&= x^{2} + 2	imes x 	imes \frac{a}{2} + \left(\frac{a}{2}ight)^{2} + b - \left(\frac{a}{2}ight)^{2}\
&= \left( x + \frac{a}{2} ight)^{2} + b - \left(\frac{a}{2}ight)^{2}
\end{align*}

The above expression will have minimum value at  x = -\frac{a}{2} .

Thanks.

jfranciscocuencag · Answer

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

ajmekal · Answer

Please remember the basic rule for quadratic equations:
ax^2 + bx + c = y
Here since y is an  upward facing parabola if a>0 we'll have minimum value -b/2a  (max value reaching +ve infinity), similarly if a<0 we'll have a  downward facing parabola with maximum value -b/2a  (min value reaches -ve infinity).

Hence the answer will be -b/2a i.e., -a/2

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If y = x2 + ax + b, y is minimum when x is:

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