\(\text{Let $a$ be a real number for which there exists a unique value of $b$ such that the quadratic equation $x^2 + 2bx + (a-b) = 0$ has one real solution. Find $a$. }\)
+128406 x^2 + 2bx + (a - b) = 0
If this has one real solution then the discriminant must = 0 ......so.....
(2b)^2 - 4(a - b) = 0
4b^2 - 4a + 4b = 0
b^2 - a + b = 0
a = b^2 + b
Note that the orginal equation reduces to
x^2 +2bx + b^2 = 0 which factors to
(x + b)^2 = 0 which only has one real solution......x = -b
Thanks, but sorry, that isn't right.
Here is the solution: file:///home/chronos/u-35de90f745c9428dae3cb17b96a0f0170e6f95e8/MyFiles/Downloads/Screenshot%202019-12-31%20at%2011.47.55.png
