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Difference between revisions of "1955 AHSME Problems/Problem 30"

(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)...")
 
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<math> \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} </math>
 
<math> \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} </math>
  
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[[1955 AHSME Problems/Problem 30|Solution]]
 
 
==Solution==
 
==Solution==
 
Since the question asks us about the unifying characteristic of all three equations' roots, we have to first determine them.
 
Since the question asks us about the unifying characteristic of all three equations' roots, we have to first determine them.
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We can clearly see that, between all of the equations, there is <math>\boxed{\textbf{(B)} \text{no root greater than 3}}</math>.
 
We can clearly see that, between all of the equations, there is <math>\boxed{\textbf{(B)} \text{no root greater than 3}}</math>.
 +
 +
Note: There are probably extraneous solutions somewhere, but that does not affect the solution.
 +
 
==See Also==
 
==See Also==
  

Latest revision as of 19:10, 28 September 2024

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.

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

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