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Joined: 03 Jul 2015
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If y = x2 + ax + b, y is minimum when x is:
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27 Dec 2015, 00:20
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58% (01:42) correct 42% (02:05) wrong based on 97 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 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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Re: If y = x2 + ax + b, y is minimum when x is:
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27 Dec 2015, 06:16
shreyashid wrote: 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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Re: If y = x2 + ax + b, y is minimum when x is:
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28 Dec 2015, 18:15
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 > 44+1 = +1 Answer C: A/2 = 2/2 = 1 > 12+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
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Re: If y = x2 + ax + b, y is minimum when x is:
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29 Dec 2015, 03:08
shreyashid wrote: 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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Joined: 20 Jun 2015
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Concentration: Finance, Human Resources

Re: If y = x2 + ax + b, y is minimum when x is:
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12 Jan 2016, 23:45
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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Joined: 03 Jul 2015
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Re: If y = x2 + ax + b, y is minimum when x is:
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13 Jan 2016, 00:28
thanks everyone. the Parabola logic makes it look so simple



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Re: If y = x2 + ax + b, y is minimum when x is:
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15 Sep 2016, 03:16
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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Re: If y = x2 + ax + b, y is minimum when x is:
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18 Feb 2017, 05:52
shreyashid wrote: 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:
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29 Mar 2019, 17:44
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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Joined: 22 Mar 2019
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Re: If y = x2 + ax + b, y is minimum when x is:
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31 Mar 2019, 20:34
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




Re: If y = x2 + ax + b, y is minimum when x is:
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31 Mar 2019, 20:34






