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For what value of a will this equation have only one real root: (2a−5)x^2−2(a−1)x+3=0.

Guest Dec 10, 2017
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(2a−5)x^2−2(a−1)x+3=0

 

Let b =  -2(a - 1)  =   2 - 2a

Let a  =  (2a - 5)

Let c  =  3

 

This will  happen when

 

b^2 -  4ac  = 0    ....so we have.....

 

(2 - 2a)^2  - 4(2a - 5)(3)   =  0     simplify

 

4a^2 - 8a + 4   -  24a  + 60  = 0

 

4a^2 - 32a + 64  =  0

 

a^2 - 8a + 16  = 0

 

(a - 4)^2  = 0

 

And a   =  4 

 

 

cool cool cool

CPhill  Dec 11, 2017

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