1967 AHSME Problems/Problem 35
Contents
Problem
The roots of 
are in arithmetic progression. The difference between the largest and smallest roots is:
Solution 1
By Vieta's Formulas, the sum of the roots is 
. Because the roots are in arithmetic progression, the middle root is the average of the other two roots, and is also the average of all three roots. Therefore, 
is the middle root.
The other two roots have a sum of 
. By Vieta on the original cubic, the product of all 
roots is 
, so the product of the remaining two roots is 
. If the sum of the two remaining roots is 
, and the product is 
, the two remaining roots are also the two roots of 
.
The two remaining roots are thus 
, and they have a difference of 
. Plugging in gives 
, which is equal to 
, which is answer 
.
Solution 2
Because the roots are in arithmetic progression, let the roots be 
and 
with 
. From Vieta's Formulas, we know that 
, so 
, and 
.
By using Vieta again, we see that 
. Substituting into this equation, we can now solve for 
:
\begin{align*}
(\frac 3 4-d)(\frac 3 4)(\frac 3 4+d) &= \frac{15}{64} \\
\frac 3 4(\frac 9{16}-d^2) &= \frac{15}{64} \\
\frac 9{16}-d^2 &= \frac 5{16} \\
d^2 &= \frac 4{16} = \frac 1 4 \\
d &= \pm \frac 1 2
\end{align*}
Because we defined , 
. Therefore, the difference between the largest and smallest roots, and 
, is 
. h
See also
| 1967 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 34 |
Followed by Problem 36 | |
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