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>
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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 $i=\sqrt{-1}$. The product of the real parts of the roots of $z^2-z=5-5i$ is

$\text{(A) } -25\quad \text{(B) } -6\quad \text{(C) } -5\quad \text{(D) } \frac{1}{4}\quad \text{(E) } 25$

Solution

Applying the quadratic formula gives \[z=\frac{-1\pm\sqrt{21-20i}}{2}\]

Let \[\sqrt{21-20i}=a+bi\] where $a$ and $b$ are real. Squaring both sides and equating real and imaginary parts gives \[a^2-b^2=21\] \[2ab=-20\] Substituting $b=-\frac{10}{a}$, letting $a^2=n$ and solving gives $n=25, -4$, from which we see that $a=5$ and $b=-2$.

Replacing the square root in the quadratic formula with $5-2i$ and simplifying gives the two roots $2-i$ and $-3+i$. The product of their real parts is $-6$. The answer is $\fbox{B}$.

See also

1992 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 27
Followed by
Problem 29
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