Difference between revisions of "2005 AMC 10A Problems/Problem 10"
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<cmath> (a+8)^2 - 4(4)(9) = 0 \implies a^2 + 16a - 80. </cmath> We can use the quadratic formula to solve for its roots (we can ignore the things in the radical sign as they will cancel out due to the <math>\pm</math> sign when added). So we must have | <cmath> (a+8)^2 - 4(4)(9) = 0 \implies a^2 + 16a - 80. </cmath> We can use the quadratic formula to solve for its roots (we can ignore the things in the radical sign as they will cancel out due to the <math>\pm</math> sign when added). So we must have | ||
<cmath> {-16 + \sqrt{\text{something}}}{2} + \frac{-16 - \sqrt{\text{something}}}{2}. </cmath> | <cmath> {-16 + \sqrt{\text{something}}}{2} + \frac{-16 - \sqrt{\text{something}}}{2}. </cmath> | ||
− | Therefore, we have \implies \boxed{A}. | + | Therefore, we have <math>\implies \boxed{A}</math>. |
== Solution 3== | == Solution 3== |
Revision as of 19:03, 17 September 2019
Contents
[hide]Problem
There are two values of for which the equation
has only one solution for
. What is the sum of those values of
?
Solution 1
A quadratic equation has exactly one root if and only if it is a perfect square. So set
Two polynomials are equal only if their coefficients are equal, so we must have
or
.
So the desired sum is
Alternatively, note that whatever the two values of are, they must lead to equations of the form
and
. So the two choices of
must make
and
so
and
.
Solution 2
Since this quadratic must have a double root, the discriminant of the quadratic formula for this quadratic must be 0. Therefore, we must have
We can use the quadratic formula to solve for its roots (we can ignore the things in the radical sign as they will cancel out due to the
sign when added). So we must have
Therefore, we have
.
Solution 3
There is only one positive value for k such that the quadratic equation would have only one solution.
k-8 and -k-8 are the values of a. I am certain we all know what -8-8 is. That’s correct,
See also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.