1981 AHSME Problems/Problem 1: Difference between revisions

Revision as of 00:26, 15 January 2020

If $\sqrt{x+2}=2$ , then $(x+2)^2$ equals:

$\textbf{(A)}\ \sqrt{2}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 16$

If we square both sides of the $\sqrt{x+2} = 2$ , we will get $x+2 = 4$ , if we square that again, we get $(x+2)^2 = \boxed{\textbf{(E) }16}$

We can immediately get that $x = 2$ , after we square $(2+2)$ , we get $\boxed{\textbf{(E) }16}$

@@ Line 5: / Line 5: @@
-==Solution==
+==Solution 1==
-<math>\boxed{\textbf{(E) }16}</math>
+If we square both sides of the <math>\sqrt{x+2} = 2</math>, we will get <math>x+2 = 4</math>, if we square that again, we get <math>(x+2)^2 = \boxed{\textbf{(E) }16}</math>
+==Solution 2==
+We can immediately get that <math>x = 2</math>, after we square <math>(2+2)</math>, we get <math>\boxed{\textbf{(E) }16}</math>