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Difference between revisions of "1970 AHSME Problems/Problem 14"
 
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== Solution ==
 
== Solution ==
−
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>.
+
From the quadratic equation, the two roots of the equation are <math>\frac{-p\pm\sqrt{p^2-4q}}{2}</math>. The positive difference between these roots is <math>\sqrt{p^2 - 4q}</math>. Setting <math>\sqrt{p^2-4q}=1</math> and isolating <math>p</math> gives <math>\sqrt{4q+1}</math>, or choice <math>\boxed{\text{(A)}}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 00:30, 24 February 2023

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 $\frac{-p\pm\sqrt{p^2-4q}}{2}$. The positive difference between these roots is $\sqrt{p^2 - 4q}$. Setting $\sqrt{p^2-4q}=1$ and isolating $p$ gives $\sqrt{4q+1}$, or choice $\boxed{\text{(A)}}$.

See also

1970 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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All AHSME Problems and Solutions

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