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

 Mar 7, 2024
 #1
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Simplify as

 

x^2 + (4m - 2)x + (m + 6)  =  0

 

We have one root when  the discriminant  =  0

 

(4m - 2)^2 - 4 (1) ( m + 6)  = 0

 

16m^2 - 16m + 4 - 4m  - 24   = 0

 

16m^2 -20m - 20  =  0 

 

4m^2 - 5m - 5  = 0 

 

m =   [ 5 + sqrt [ 25 + 80 ] ] / 8  =    [ 5 + sqrt 105 ] / 8

 

 

cool cool cool

 Mar 7, 2024

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