Difference between revisions of "1957 AHSME Problems/Problem 14"

Latest revision as of 08:21, 25 July 2024

If $y = \sqrt{x^2 - 2x + 1} + \sqrt{x^2 + 2x + 1}$ , then $y$ is:

$\textbf{(A)}\ 2x\qquad \textbf{(B)}\ 2(x+1)\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ |x-1|+|x+1|\qquad\textbf{(E)}\ \text{none of these}$

Solution

In order to solve the problem, we will use two properties, namely, that $(a+b)^2=a^2+2ab+b^2$ and $(a-b)^2=a^2-2ab+b^2$ .

We can use this to simplify the equation, as $y = \sqrt{x^2 - 2x + 1} + \sqrt{x^2 + 2x + 1}$ turns into $y = x + 1 + x - 1$ . However, the square root function only allows for nonnegative inputs and only generates nonnegative outputs, so $y$ is $\boxed{\textbf{(D) }|x+1|+|x-1|}$ .

1957 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 8: / Line 8: @@
 ==See Also==
-{{AHSME box|year=1957|num-b=13|num-a=15}}
+{{AHSME 50p box|year=1957|num-b=13|num-a=15}}
 {{MAA Notice}}
 [[Category:AHSME]][[Category:AHSME Problems]]

Page

Toolbox

Search

Difference between revisions of "1957 AHSME Problems/Problem 14"

Latest revision as of 08:21, 25 July 2024

Solution

See Also