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!
+128474 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
