https://artofproblemsolving.com/wiki/index.php?title=1978_AHSME_Problems/Problem_13&feed=atom&action=history
1978 AHSME Problems/Problem 13 - Revision history
2024-03-29T05:48:11Z
Revision history for this page on the wiki
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Coolmath34 at 01:06, 14 February 2021
2021-02-14T01:06:27Z
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 01:06, 14 February 2021</td>
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<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">== Problem 13 ==</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">If <math>a,b,c</math>, and <math>d</math> are non-zero numbers such that <math>c</math> and <math>d</math> are the solutions of <math>x^2+ax+b=0</math> and <math>a</math> and <math>b</math> are </ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">the solutions of <math>x^2+cx+d=0</math>, then <math>a+b+c+d</math> equals</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"><math>\textbf{(A) }0\qquad</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">\textbf{(B) }-2\qquad</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">\textbf{(C) }2\qquad</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">\textbf{(D) }4\qquad </ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">\textbf{(E) }(-1+\sqrt{5})/2  </math> </ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">== Solution ==</ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>By Vieta's formulas, <math>c + d = -a</math>, <math>cd = b</math>, <math>a + b = -c</math>, and <math>ab = d</math>. From the equation <math>c + d = -a</math>, <math>d = -a - c</math>, and from the equation <math>a + b = -c</math>, <math>b = -a - c</math>, so <math>b = d</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>By Vieta's formulas, <math>c + d = -a</math>, <math>cd = b</math>, <math>a + b = -c</math>, and <math>ab = d</math>. From the equation <math>c + d = -a</math>, <math>d = -a - c</math>, and from the equation <math>a + b = -c</math>, <math>b = -a - c</math>, so <math>b = d</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Then from the equation <math>cd = b</math>, <math>cb = b</math>. Since <math>b</math> is nonzero, we can divide both sides of the equation by <math>b</math> to get <math>c = 1</math>. Similarly, from the equation <math>ab = d</math>, <math>ab = b</math>, so <math>a = 1</math>. Then <math>b = d = -a - c = -2</math>. Therefore, <math>a + b + c + d = 1 + (-2) + 1 + (-2) = \boxed{-2}</math>. The answer is (B).</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Then from the equation <math>cd = b</math>, <math>cb = b</math>. Since <math>b</math> is nonzero, we can divide both sides of the equation by <math>b</math> to get <math>c = 1</math>. Similarly, from the equation <math>ab = d</math>, <math>ab = b</math>, so <math>a = 1</math>. Then <math>b = d = -a - c = -2</math>. Therefore, <math>a + b + c + d = 1 + (-2) + 1 + (-2) = \boxed{-2}</math>. The answer is (B).</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">==See Also==</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">{{AHSME box|year=1978|num-b=12|num-a=14}}</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">{{MAA Notice}}</ins></div></td></tr>
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Coolmath34
https://artofproblemsolving.com/wiki/index.php?title=1978_AHSME_Problems/Problem_13&diff=94919&oldid=prev
Apple2017: Answer is (B): -2
2018-06-05T00:46:41Z
<p>Answer is (B): -2</p>
<p><b>New page</b></p><div>By Vieta's formulas, <math>c + d = -a</math>, <math>cd = b</math>, <math>a + b = -c</math>, and <math>ab = d</math>. From the equation <math>c + d = -a</math>, <math>d = -a - c</math>, and from the equation <math>a + b = -c</math>, <math>b = -a - c</math>, so <math>b = d</math>.<br />
<br />
Then from the equation <math>cd = b</math>, <math>cb = b</math>. Since <math>b</math> is nonzero, we can divide both sides of the equation by <math>b</math> to get <math>c = 1</math>. Similarly, from the equation <math>ab = d</math>, <math>ab = b</math>, so <math>a = 1</math>. Then <math>b = d = -a - c = -2</math>. Therefore, <math>a + b + c + d = 1 + (-2) + 1 + (-2) = \boxed{-2}</math>. The answer is (B).</div>
Apple2017