1955 AHSME Problems/Problem 30

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Problem 30

Each of the equations $3x^2-2=25, (2x-1)^2=(x-1)^2, \sqrt{x^2-7}=\sqrt{x-1}$ has:

$\textbf{(A)}\ \text{two integral roots}\qquad\textbf{(B)}\ \text{no root greater than 3}\qquad\textbf{(C)}\ \text{no root zero}\\ \textbf{(D)}\ \text{only one root}\qquad\textbf{(E)}\ \text{one negative root and one positive root}$

Solution

Since the question asks us about the unifying characteristic of all three equations' roots, we have to first determine them.

$3x^2-2 = 25$ can be rewritten as $3x^2 - 27 = 0$, which gives the following roots $+3$ and $-3$.

$(2x-1)^2 = (x-1)^2$ can be expanded to $4x^2-4x+1=x^2-2x+1$, which in turn leads to $3x^2-2x=0$. The roots here are $0$ and $\frac{2}{3}$.

$\sqrt{x^2-7}=\sqrt{x-1}$, when squared, also turns into a quadratic equation: $x^2 - x - 6 = 0$. Binomial factoring gives us the roots $-2$ and $3$.

We can clearly see that, between all of the equations, there is $\boxed{\textbf{(B)} \text{no root greater than 3}}$.

Note: There are probably extraneous solutions somewhere, but that does not affect the solution.

See Also

In order to go back to the 1955 AHSME, click here.

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