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The quadratic equation ax^2 + 8x + x = 0 has exactly one solution. If a + c = 10, and a < c, find the ordered pair (a, c).

 

My original answer was (-8, -2), but it was incorrect so I must have done something wrong.

 

Thanks in advance.

Guest Feb 13, 2018
 #1
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Do you mean this ???

 

ax^2 + 8x + c   =  0    ???   if so

 

If we have only one solution, the discriminant will = 0   .....so.....

 

8^2  - 4ac  = 0            and      a + c  =  10   ⇒    c  =  10 - a

 

So....substituting, we have

 

8^2 - 4a(10 - a)   = 0

 

8^2  - 40a + 4a^2  =  0

 

4a^2 - 40a +  64  =  0         divide through by 4

 

a^2  - 10a  + 16   =  0      factor

 

(a - 8) ( a - 2)  =  0

 

The solutions are a  = 8    or a  = 2

 

Note that if a  = 8, then c  = 2     or

If a  = 2  then c  = 8

 

But a > c  so     (a, c)   =   (  8, 2 )

 

Check the ggraph here that only one zero exists  at  x  = -1/2 :

 

https://www.desmos.com/calculator/idsuikzats

 

 

cool cool cool

CPhill  Feb 13, 2018

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