Difference between revisions of "1970 AHSME Problems/Problem 14"

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== Solution ==
 
== Solution ==
<math>\fbox{A}</math>
+
From the quadratic equation, the two roots of the equation are <math>\frac{-p + \sqrt{p^2 - 4q}}{2}</math> and <math>\frac{-p - \sqrt{p^2 - 4q}}{2}</math>.  The positive difference between these roots is <math>\sqrt{p^2 - 4q}</math>.  If <math>\sqrt{p^2 - 4q} = 1</math>, then <math>p^2 - 4q = 1</math>.  This leads to <math>p^2 = 1 + 4q</math>, and <math>p = \sqrt{1 + 4q}</math>, which is answer <math>\fbox{A}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 21:18, 13 July 2019

Problem

Consider $x^2+px+q=0$, where $p$ and $q$ are positive numbers. If the roots of this equation differ by 1, then $p$ equals

$\text{(A) } \sqrt{4q+1}\quad \text{(B) } q-1\quad \text{(C) } -\sqrt{4q+1}\quad \text{(D) } q+1\quad \text{(E) } \sqrt{4q-1}$

Solution

From the quadratic equation, the two roots of the equation are $\frac{-p + \sqrt{p^2 - 4q}}{2}$ and $\frac{-p - \sqrt{p^2 - 4q}}{2}$. The positive difference between these roots is $\sqrt{p^2 - 4q}$. If $\sqrt{p^2 - 4q} = 1$, then $p^2 - 4q = 1$. This leads to $p^2 = 1 + 4q$, and $p = \sqrt{1 + 4q}$, which is answer $\fbox{A}$.

See also

1970 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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