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Find the constant k such that the quadratic 2x^2 + 3x + 8x - x^2 + 4x + k has a double root.

 Jul 4, 2024
 #1
avatar+118 
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for a quadratic to have a double root we want the variables of b^2 -4ac = 0

 

we can combine like terms in this quadratic via simplification and get 

 

x^2+11x+4x + k 

==> x^2 + 15x + k = 0

now we just plug this into the discriminant equation ( b^2 - 4ac = 0  ) 

== > 15^2 - 4 x 1 x c =0

simplfy == > 225 - 4c = 0 

thus c = 56.25

 

please check if this is right 

 Jul 4, 2024
 #3
avatar+944 
+1

 

Yes, it is correct.  I was still typing when your answer was   

posted, but we both solved it the same way.  Good work.   

,

Bosco  Jul 4, 2024
 #2
avatar+944 
+1

 

Find the constant k such that the quadratic 2x^2 + 3x + 8x - x^2 + 4x + k has a double root.  

 

                                                                    2x2 + 3x + 8x – x2 + 4x + k   

 

combine like terms                                         x2 + 15x + k   

 

for a quadratic to have a double root,   

which is also called a repeated root,   

its discriminant must equal zero

 

when a quadratic is in the form ax2 + bx +c   

its discriminant is the quantity b2 – 4ac            b2 – 4ac   

                                                                       152 – (4)(1)(k)   

                                                                       225 – 4k    

 

solve for k                                                          k  =  (225 / 4)    

 

                                                                           k  =  56.25    

 

by the way, the quadratic in this problem factors to (x + 7.5)(x + 7.5)   

.

 Jul 4, 2024

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