1955 AHSME Problems/Problem 30

Revision as of 15:01, 11 August 2020 by Angrybird029 (talk | contribs) (Created page with "== Problem 30== Each of the equations <math>3x^2-2=25, (2x-1)^2=(x-1)^2, \sqrt{x^2-7}=\sqrt{x-1}</math> has: <math> \textbf{(A)}\ \text{two integral roots}\qquad\textbf{(B)...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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

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 $\boxed{\textbf{(B)} \text{no root greater than 3}}$.

See Also

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

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