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Difference between revisions of "1965 AHSME Problems/Problem 1"
m (fixed boxed answer choice to correspond to derived/correct answer)
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==Solution2==
 
==Solution2==
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Notice that <math>a^0=1, a>0</math>. So <math>2^0=1</math>. So <math>2x^2-7x+5=0</math>. Evaluating the discriminant, we see that it is equal to <math>7^2-4*2*5=9</math>. So this means that the equation has two real solutions. Therefore, select <math>\boxed{B}</math>.
+
Notice that <math>a^0=1, a>0</math>. So <math>2^0=1</math>. So <math>2x^2-7x+5=0</math>. Evaluating the discriminant, we see that it is equal to <math>7^2-4*2*5=9</math>. So this means that the equation has two real solutions. Therefore, select <math>\fbox{C}</math>.
  
 
~hastapasta
 
~hastapasta

Revision as of 08:09, 18 July 2024

Problem

The number of real values of $x$ satisfying the equation $2^{2x^2 - 7x + 5} = 1$ is:

$\textbf{(A)}\ 0 \qquad \textbf{(B) }\ 1 \qquad \textbf{(C) }\ 2 \qquad \textbf{(D) }\ 3 \qquad \textbf{(E) }\ \text{more than 4}$


Solution 1

Solution by e_power_pi_times_i

Take the logarithm with a base of $2$ to both sides, resulting in the equation $2x^2-7x+5 = 0$. Factoring results in $(2x-5)(x-1) = 0$, so there are $\boxed{\textbf{(C) } 2}$ real solutions.

Solution2

Notice that $a^0=1, a>0$. So $2^0=1$. So $2x^2-7x+5=0$. Evaluating the discriminant, we see that it is equal to $7^2-4*2*5=9$. So this means that the equation has two real solutions. Therefore, select $\fbox{C}$.

~hastapasta

See Also

1965 AHSME (Problems • Answer Key • Resources)
Preceded by
First Question 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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