+0  
 
0
42
1
avatar

Let a be a real number for which there exists a unique value of b such that the quadratic equation x^2 + 2bx + (a-b) = 0 has one real solution. Find a.

 

Hello Cphill I have tried asking this many times but have not gotten a clear answer! It would help if you Said your answer clearly, like saying “a equals this”. Thank you very much for your help, you have been very helpful to me in the past!

 Mar 27, 2020
 #1
avatar+111330 
+1

Gotta' give  credit to Geno for his solution here :

 

https://web2.0calc.com/questions/answer-was-not-clearly-stated-last-time

 

His answer  is all that we can infer  because  the numerical value of a  depends upon  the  numerical value of b

 

For instance....pick a  value  for  b, say ,1

 

Then a =  1^2 + 1   = 2

 

Then  we have  that 

 

x^2  + 2(1)x + (2 - 1)  = 0

 

x^2 + 2x  + 1   =  0

 

(x + 1)^2   =    0

 

Verify for yourself that  x = -1   will be the  only  solution....

 

 

Also  note  that, in general,  we have

 

 x^2  + 2bx  + (b^2 + b - b)   =  0

 

x^2  + 2bx  +  b^2   =  0      which  can be factored as

 

(x + b)^2  =  0

 

So....we will always  have  just one solution, namely......x  = -b

 

 

 

 

 

cool cool cool

 Mar 27, 2020
edited by CPhill  Mar 27, 2020
edited by CPhill  Mar 27, 2020

11 Online Users