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.
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I plan to use the quadratic formula on this equation: x2 + 2bx + (a - b) = 0
using a = 1 b = 2b c = a - b
For the equation to have only one real solution b2 - 4ac must be zero.
Replacing: (2b)2 - 4(1)(a - b) = 0
4b2 - 4a + 4b = 0
-4a = -4b2 - 4b
Diviing by -4: a = b2 + b
I hope that the variables in the equation did not confuse you with the variables in the quadratic formula.
Hint:
One root means that the equation has to be in the form of (x+#)^2=0.
Hope this helped!
I plan to use the quadratic formula on this equation: x2 + 2bx + (a - b) = 0
using a = 1 b = 2b c = a - b
For the equation to have only one real solution b2 - 4ac must be zero.
Replacing: (2b)2 - 4(1)(a - b) = 0
4b2 - 4a + 4b = 0
-4a = -4b2 - 4b
Diviing by -4: a = b2 + b
I hope that the variables in the equation did not confuse you with the variables in the quadratic formula.