Difference between revisions of "2005 AMC 10A Problems/Problem 10"
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There is only one positive value for <math>k</math> such that the quadratic equation would have only one solution. | There is only one positive value for <math>k</math> such that the quadratic equation would have only one solution. | ||
<math>k-8</math> and <math>-k-8</math> are the values of <math>a</math>. <math>-8-8=-16</math>, so the answer is <math>\boxed{\textbf{(A)} -16}</math> | <math>k-8</math> and <math>-k-8</math> are the values of <math>a</math>. <math>-8-8=-16</math>, so the answer is <math>\boxed{\textbf{(A)} -16}</math> | ||
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+ | Another way of thinking of this is letting the two values of k be <math>a</math> and <math>a_0</math>. Since this is obviously a square of a binomial, <math>k+8 = m</math> or <math>-m</math> for some <math>m</math>. Thus, we can say that <math>a + 8 = m</math> and <math>a_0 + 8 = -m</math>. Combining these gives us <math>a_0 + a + 16 = 0</math>, so <math>a_0 + a = 16</math>. Our answer is <math>\boxed{\textbf{(A)} -16}</math>. | ||
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+ | ~Extremelysupercooldude (Second solution) | ||
==See also== | ==See also== |
Latest revision as of 18:05, 29 March 2023
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 expression in the radical sign as it will cancel out due to the
sign when added). So we must have
Solution 3
There is only one positive value for such that the quadratic equation would have only one solution.
and
are the values of
.
, so the answer is
Another way of thinking of this is letting the two values of k be and
. Since this is obviously a square of a binomial,
or
for some
. Thus, we can say that
and
. Combining these gives us
, so
. Our answer is
.
~Extremelysupercooldude (Second solution)
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 |
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