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For what value of the constant k\(\) does the quadratic \( 2x^2 - 5x + k\) have a double root?

 Dec 27, 2017
edited by Guest  Dec 27, 2017

Best Answer 

 #3
avatar+2339 
+2

Rahuan, I am not completely sure what it means either, but I will interpret "double root" as a zero with a multiplicity of two. Assuming the above is true, we can use the property of the discriminant of a quadratic, \(b^2-4ac\) . The discriminant, if equal to zero, results in a quadratic with a "double root." 

 

\(b^2-4ac=0\) Plug in the given values of a, b, and c of the quadratic. 
\((-5)^2-4(2)(k)=0\) Now, solve for k. 
\(25-8k=0\) Subtract 25 from both sides. 
\(-8k=-25\)  
\(k=\frac{25}{8} \)  
   
   
   
   
   
   
   
   
   
   
 Dec 27, 2017
 #1
avatar+502 
0

I understood until the equation part but what do u mean by double root?

 Dec 27, 2017
 #2
avatar+502 
0

or do u mean square root/ just (x2). (x = any number/ coefficients etc)

Rauhan  Dec 27, 2017
 #5
avatar
0

a repeated root

Guest Dec 27, 2017
 #3
avatar+2339 
+2
Best Answer

Rahuan, I am not completely sure what it means either, but I will interpret "double root" as a zero with a multiplicity of two. Assuming the above is true, we can use the property of the discriminant of a quadratic, \(b^2-4ac\) . The discriminant, if equal to zero, results in a quadratic with a "double root." 

 

\(b^2-4ac=0\) Plug in the given values of a, b, and c of the quadratic. 
\((-5)^2-4(2)(k)=0\) Now, solve for k. 
\(25-8k=0\) Subtract 25 from both sides. 
\(-8k=-25\)  
\(k=\frac{25}{8} \)  
   
   
   
   
   
   
   
   
   
   
TheXSquaredFactor Dec 27, 2017
 #4
avatar+502 
0

does he mean like finding the value of k and then putting the answer as \(\frac{5^2}{8}\)

Rauhan  Dec 27, 2017
 #6
avatar+2339 
0

I do not believe so; there is no reason not to simplify the fraction completely. 

TheXSquaredFactor  Dec 28, 2017

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