Difference between revisions of "1954 AHSME Problems/Problem 24"

(Created page with "== Problem 24== The values of <math>k</math> for which the equation <math>2x^2-kx+x+8=0</math> will have real and equal roots are: <math> \textbf{(A)}\ 9\text{ and }-7\qqua...")
 
 
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== Solution ==
 
== Solution ==
 
<math>b^2-4ac=0\implies b^2=4ac\implies (1-k)^2=4(2)(8)\implies (1-k)^2=8^2\implies 1-k=8\implies k=-7, 1-k=-8\implies k=9</math>, <math>\fbox{A}</math>
 
<math>b^2-4ac=0\implies b^2=4ac\implies (1-k)^2=4(2)(8)\implies (1-k)^2=8^2\implies 1-k=8\implies k=-7, 1-k=-8\implies k=9</math>, <math>\fbox{A}</math>
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==See Also==
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{{AHSME 50p box|year=1954|num-b=23|num-a=25}}
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{{MAA Notice}}

Latest revision as of 00:29, 28 February 2020

Problem 24

The values of $k$ for which the equation $2x^2-kx+x+8=0$ will have real and equal roots are:

$\textbf{(A)}\ 9\text{ and }-7\qquad\textbf{(B)}\ \text{only }-7\qquad\textbf{(C)}\ \text{9 and 7}\\ \textbf{(D)}\ -9\text{ and }-7\qquad\textbf{(E)}\ \text{only 9}$

Solution

$b^2-4ac=0\implies b^2=4ac\implies (1-k)^2=4(2)(8)\implies (1-k)^2=8^2\implies 1-k=8\implies k=-7, 1-k=-8\implies k=9$, $\fbox{A}$

See Also

1954 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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All AHSME Problems and Solutions


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