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I need help with quadratic equations!!

 

The quadratic equation $x^2+4mx+m = 2x - 6$ has exactly one real root.  Find the positive value of $m$.

 Sep 7, 2023
 #1
avatar+126978 
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Rearrange as     x^2 + (4m -2) + (m + 6)   = 0

 

If we have one real root....the discriminant =  0 

 

So

 

(4m - 2)^2  - 4 (1) ( m + 6) = 0

 

16m^2 - 16m + 4 - 4m - 24  = 0

 

16m^2 - 20m - 20  =   0      divide through by 4

 

4m^2  - 5m - 5 =  0

 

 

m =  [ 5 + sqrt [ 25 - 4 (4)(-5) ] ] / (2 * 4 )  =   [  5 + sqrt ( 105)   ]  /  8

 

cool cool cool

 Sep 7, 2023

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