Difference between revisions of "2020 AMC 10A Problems/Problem 5"
MRENTHUSIASM (talk | contribs) m (→Solution 3 (Casework and Graphing)) |
Lotusjayden (talk | contribs) (→Solution 1 (Casework and Factoring): (there was a extra negative sign in x^2-12x+34=-2. I believe it is a typo.)) |
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Case 2: | Case 2: | ||
− | Similarly, taking the nonpositive case for the value inside the absolute value notation yields <math> | + | Similarly, taking the nonpositive case for the value inside the absolute value notation yields <math>x^2-12x+34=-2</math>. Factoring and simplifying gives <math>(x-6)^2=0</math>, so the only value for this case is <math>6</math>. |
Summing all the values results in <math>4+8+6=\boxed{\textbf{(C) }18}</math>. | Summing all the values results in <math>4+8+6=\boxed{\textbf{(C) }18}</math>. | ||
Line 33: | Line 33: | ||
<ol style="margin-left: 1.5em;"> | <ol style="margin-left: 1.5em;"> | ||
<li><math>(x-6)^2-2=2</math></li><p> | <li><math>(x-6)^2-2=2</math></li><p> | ||
− | The line <math>y=2</math> intersects the parabola <math>y=(x-6)^2-2</math> at two points | + | The line <math>y=2</math> intersects the parabola <math>y=(x-6)^2-2</math> at two points that are symmetric about the line <math>x=6.</math><p> |
In this case, the average of the solutions is <math>6,</math> so the sum of the solutions is <math>12.</math> | In this case, the average of the solutions is <math>6,</math> so the sum of the solutions is <math>12.</math> | ||
<li><math>(x-6)^2-2=-2</math></li><p> | <li><math>(x-6)^2-2=-2</math></li><p> | ||
− | The line <math>y=-2</math> intersects the parabola <math>y=(x-6)^2-2</math> at one point | + | The line <math>y=-2</math> intersects the parabola <math>y=(x-6)^2-2</math> at one point: the vertex of the parabola.<p> |
In this case, the only solution is <math>x=6.</math> | In this case, the only solution is <math>x=6.</math> | ||
</ol> | </ol> |
Latest revision as of 18:38, 12 December 2022
Contents
Problem
What is the sum of all real numbers for which
Solution 1 (Casework and Factoring)
Split the equation into two cases, where the value inside the absolute value is positive and nonpositive.
Case 1:
The equation yields , which is equal to
. Therefore, the two values for the positive case is
and
.
Case 2:
Similarly, taking the nonpositive case for the value inside the absolute value notation yields . Factoring and simplifying gives
, so the only value for this case is
.
Summing all the values results in .
Solution 2 (Casework and Vieta)
We have the equations and
.
Notice that the second is a perfect square with a double root at , and the first has two distinct real roots. By Vieta's, the sum of the roots of the first equation is
or
.
.
Solution 3 (Casework and Graphing)
Completing the square gives
Note that the graph of
is an upward parabola with the vertex
and the axis of symmetry
the graphs of
are horizontal lines.
We apply casework to
The line intersects the parabola
at two points that are symmetric about the line
In this case, the average of the solutions is so the sum of the solutions is
The line intersects the parabola
at one point: the vertex of the parabola.
In this case, the only solution is
Finally, the sum of all solutions is
~MRENTHUSIASM
Video Solution 1
Video Solution 2
Education, The Study Of Everything
~IceMatrix
Video Solution 3
https://www.youtube.com/watch?v=7-3sl1pSojc
~bobthefam
Video Solution 4
~savannahsolver
Video Solution 5
https://youtu.be/3dfbWzOfJAI?t=1544
~ pi_is_3.14
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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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