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605-655 Level|   Algebra|      
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shreyashid
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If y = x2 + ax + b, y is minimum when x is: 27 Dec 2015, 00:20
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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45% (medium)

Question Stats:

63% (01:38) correct 37% (02:05) wrong based on 212 sessions

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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

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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?­
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C
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chetan2u
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If y = x2 + ax + b, y is minimum when x is: 27 Dec 2015, 06:16
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shreyashid
If y = x2 + 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,
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..
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If y = x2 + ax + b, y is minimum when x is: 28 Dec 2015, 18:15
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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:
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C

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If y = x2 + ax + b, y is minimum when x is: 29 Dec 2015, 03:08
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shreyashid
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?

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
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If y = x2 + ax + b, y is minimum when x is: 12 Jan 2016, 23:45
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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.
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If y = x2 + ax + b, y is minimum when x is: 13 Jan 2016, 00:28
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thanks everyone. the Parabola logic makes it look so simple :)
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sanaexam
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If y = x2 + ax + b, y is minimum when x is: 15 Sep 2016, 03:16
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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
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If y = x2 + ax + b, y is minimum when x is: 18 Feb 2017, 05:52
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shreyashid
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,

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

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

Thanks.
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If y = x2 + ax + b, y is minimum when x is: 29 Mar 2019, 17:44
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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
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If y = x2 + ax + b, y is minimum when x is: 31 Mar 2019, 20:34
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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: 22 Jan 2025, 07:14
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