Difference between revisions of "1992 AHSME Problems/Problem 28"
(Created page with "== Problem == Let <math>i=\sqrt{-1}</math>. The product of the real parts of the roots of <math>z^2-z=5-5i</math> is <math>\text{(A) } -25\quad \text{(B) } -6\quad \text{(C) } ...") |
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== Solution == | == Solution == | ||
− | <math>\fbox{B}</math> | + | Applying the quadratic formula gives <cmath>z=\frac{-1\pm\sqrt{21-20i}}{2}</cmath> |
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+ | Let <cmath>\sqrt{21-20i}=a+bi</cmath> where <math>a</math> and <math>b</math> are real. Squaring both sides and equating real and imaginary parts gives <cmath>a^2-b^2=21</cmath> <cmath>2ab=-20</cmath> Substituting <math>b=-\frac{10}{a}</math>, letting <math>a^2=n</math> and solving gives <math>n=25, -4</math>, from which we see that <math>a=5</math> and <math>b=-2</math>. | ||
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+ | Replacing the square root in the quadratic formula with <math>5-2i</math> and simplifying gives the two roots <math>2-i</math> and <math>-3+i</math>. The product of their real parts is <math>-6</math>. The answer is <math>\fbox{B}</math>. | ||
== See also == | == See also == |
Latest revision as of 09:14, 1 August 2016
Problem
Let . The product of the real parts of the roots of is
Solution
Applying the quadratic formula gives
Let where and are real. Squaring both sides and equating real and imaginary parts gives Substituting , letting and solving gives , from which we see that and .
Replacing the square root in the quadratic formula with and simplifying gives the two roots and . The product of their real parts is . The answer is .
See also
1992 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
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