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Difference between revisions of "1964 AHSME Problems/Problem 24"

(Created page with "== Problem == Let <math>y=(x-a)^2+(x-b)^2, a, b</math> constants. For what value of <math>x</math> is <math>y</math> a minimum? <math>\textbf{(A)}\ \frac{a+b}{2} \qquad \tex...")
 
(→‎Solution 1)
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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.
  
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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>.
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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==

Revision as of 02:18, 24 July 2019

Problem

Let $y=(x-a)^2+(x-b)^2, a, b$ constants. For what value of $x$ is $y$ a minimum?

$\textbf{(A)}\ \frac{a+b}{2} \qquad \textbf{(B)}\ a+b \qquad \textbf{(C)}\ \sqrt{ab} \qquad \textbf{(D)}\ \sqrt{\frac{a^2+b^2}{2}}\qquad \textbf{(E)}\ \frac{a+b}{2ab}$

Solution 1

Expanding the quadratic and collecting terms gives $y = 2x^2 - (2a + 2b)x + (a^2+b^2)$. For a quadratic of the form $y = Ax^2 + Bx + C$ with $A>0$, $y$ is minimized when $x = -\frac{B}{2A}$, which is the average of the roots.

Thus, the quadratic is minimized when $x = \frac{2a+2b}{2\cdot 2} = \frac{a+b}{2}$, which is answer $\boxed{\textbf{(A)}}$.

Solution 2

The problem should return real values for $ab = 0$ and $ab < 0$, which eliminates $E$ and $C$. We want to distinguish between options $A, B, D$, and testing $(a, b) = (2, 0)$ should do that, as answers $A, B, D$ will turn into $2, 1, \sqrt{2}$, respectively.

PLugging in $(a, b) = (2, 0)$ gives $y = (x - 2)^2 + x^2$, or $y = 2x^2 - 4x + 4$. This has a minimum at $x = -\frac{-4}{2}$, or at $x=2$. This is answer $\boxed{\textbf{(A)}}$.

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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