Difference between revisions of "1964 AHSME Problems/Problem 24"
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Expanding the quadratic and collecting terms gives <math>y = 2x^2 - (2a + 2b)x + (a^2+b^2)</math>. For a quadratic of the form <math>y = Ax^2 + Bx + C</math> with <math>A>0</math>, <math>y</math> is minimized when <math>x = -\frac{B}{2A}</math>, which is the average of the roots. | Expanding the quadratic and collecting terms gives <math>y = 2x^2 - (2a + 2b)x + (a^2+b^2)</math>. For a quadratic of the form <math>y = Ax^2 + Bx + C</math> with <math>A>0</math>, <math>y</math> is minimized when <math>x = -\frac{B}{2A}</math>, which is the average of the roots. | ||
− | Thus, the quadratic is minimized when <math>x = \frac{2a+2b}{2} = \frac{a+b}{2}</math>, which is answer <math>\boxed{\textbf{(A)}}</math>. | + | Thus, the quadratic is minimized when <math>x = \frac{2a+2b}{2\cdot 2} = \frac{a+b}{2}</math>, which is answer <math>\boxed{\textbf{(A)}}</math>. |
==Solution 2== | ==Solution 2== | ||
− | The problem should return real values for <math>ab = 0</math> and <math>ab < 0</math>, which eliminates <math>E</math> and <math>C</math>. We want to distinguish between options <math>A, B, D</math>, and testing <math>(a, b) = (2, 0)</math> should do that, as answers <math>A, B, D</math> will turn into <math>2 | + | The problem should return real values for <math>ab = 0</math> and <math>ab < 0</math>, which eliminates <math>E</math> and <math>C</math>. We want to distinguish between options <math>A, B, D</math>, and testing <math>(a, b) = (2, 0)</math> should do that, as answers <math>A, B, D</math> will turn into <math>1, 2, \sqrt{2}</math>, respectively. |
− | PLugging in <math>(a, b) = (2, 0)</math> gives <math>y = (x - 2)^2 + x^2</math>, or <math>y = 2x^2 - 4x + 4</math>. This has a minimum at <math>x = -\frac{-4}{ | + | PLugging in <math>(a, b) = (2, 0)</math> gives <math>y = (x - 2)^2 + x^2</math>, or <math>y = 2x^2 - 4x + 4</math>. This has a minimum at <math>x = -\frac{-4}{4}</math>, or at <math>x=1</math>. This is answer <math>\boxed{\textbf{(A)}}</math>. |
==See Also== | ==See Also== |
Latest revision as of 03:19, 24 July 2019
Contents
Problem
Let constants. For what value of
is
a minimum?
Solution 1
Expanding the quadratic and collecting terms gives . For a quadratic of the form
with
,
is minimized when
, which is the average of the roots.
Thus, the quadratic is minimized when , which is answer
.
Solution 2
The problem should return real values for and
, which eliminates
and
. We want to distinguish between options
, and testing
should do that, as answers
will turn into
, respectively.
PLugging in gives
, or
. This has a minimum at
, or at
. This is answer
.
See Also
1964 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |
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