Difference between revisions of "1967 AHSME Problems/Problem 17"

(Created page with "== Problem == If <math>r_1</math> and <math>r_2</math> are the distinct real roots of <math>x^2+px+8=0</math>, then it must follow that: <math>\textbf{(A)}\ |r_1+r_2|>4\sqrt{2}\...")
 
 
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== Solution ==
 
== Solution ==
<math>\fbox{A}</math>
+
We are given that the roots are real, so the discriminant is positive, which means <math>p^2 - 4(8)(1) > 0</math>.  This leads to <math>|p| > 4\sqrt{2}</math>.  By Vieta, the sum of the roots is <math>-p</math>, so we have <math>|-(r_1 + r_2)| \ge 4\sqrt{2}</math>, or <math>|r_1 + r_2| > 4\sqrt{2}</math>, which is option <math>\fbox{A}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 02:03, 13 July 2019

Problem

If $r_1$ and $r_2$ are the distinct real roots of $x^2+px+8=0$, then it must follow that:

$\textbf{(A)}\ |r_1+r_2|>4\sqrt{2}\qquad \textbf{(B)}\ |r_1|>3 \; \text{or} \; |r_2| >3 \\ \textbf{(C)}\ |r_1|>2 \; \text{and} \; |r_2|>2\qquad \textbf{(D)}\ r_1<0 \; \text{and} \; r_2<0\qquad \textbf{(E)}\ |r_1+r_2|<4\sqrt{2}$

Solution

We are given that the roots are real, so the discriminant is positive, which means $p^2 - 4(8)(1) > 0$. This leads to $|p| > 4\sqrt{2}$. By Vieta, the sum of the roots is $-p$, so we have $|-(r_1 + r_2)| \ge 4\sqrt{2}$, or $|r_1 + r_2| > 4\sqrt{2}$, which is option $\fbox{A}$.

See also

1967 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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