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\(\text{Let $a$ be a real number for which there exists a unique value of $b$ such that the quadratic equation $x^2 + 2bx + (a-b) = 0$ has one real solution. Find $a$. }\)

 Dec 31, 2019
 #1
avatar+109563 
+2

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

 

 

cool cool cool

 Dec 31, 2019
edited by CPhill  Dec 31, 2019
 #2
 #3
avatar+109563 
+1

If we let x  =   - b.......we have that

 

(-b)^2  + 2b(-b)  +  ( a  - b)    = 

 

b^2  - 2b^2  +  (b^2 + b - b)   =

 

b^2  - 2b^2   +  b^2   =

 

2b^2  - 2b^2  =

 

0

 

 

cool cool cool

CPhill  Dec 31, 2019

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