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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
avatar+5652 
+1

 

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

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 Jul 16, 2019
edited by Rom  Jul 16, 2019
 #2
avatar+102386 
+1

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

 

 

cool cool cool

 Jul 16, 2019

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