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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 $$a$$ .

Jul 16, 2019

#1
+6192
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$$\text{one real solution means that the discriminant is zero}\\ \text{in the quadratic equation a x^2 + b x +c the discriminant is given by D=b^2-4ac}\\ \text{here}\\ a=1,~b=2b,~c = a-b\\ D=(2b)^2 - 4(a-b) = 4b^2 - 4a+4ab\\~\\ D=0 \Rightarrow\\ 4b^2-4a+4ab=0\\ a = b+b^2$$

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Jul 16, 2019
edited by Rom  Jul 16, 2019
#2
+111393
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To have one real solution the discriminant must = 0   ....so...

(2b)^2 - 4 (1) (a - b)  = 0

4b^2 - 4a + 4b  =  0

b^2 - a + b  = 0

b^2 + b  =   a

Jul 16, 2019