1954 AHSME Problems/Problem 21

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Problem 21

The roots of the equation $2\sqrt {x} + 2x^{ - \frac {1}{2}} = 5$ can be found by solving:

$\textbf{(A)}\ 16x^2-92x+1 = 0\qquad\textbf{(B)}\ 4x^2-25x+4 = 0\qquad\textbf{(C)}\ 4x^2-17x+4 = 0\\ \textbf{(D)}\ 2x^2-21x+2 = 0\qquad\textbf{(E)}\ 4x^2-25x-4 = 0$

Solution

Make the substitution $t=\sqrt{x}$. Then $2t+\frac{2}{t}=5\implies 2t^2-5t+2=0$. Then $2x-5\sqrt{x}+2=0\implies 2x+2=5\sqrt{x}\implies (2x+2)^2=25x\implies 4x^2+8x+4=25x\implies 4x^2-17x+4=0$, $\fbox{C}$

See Also

1954 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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