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The quadratic equation $ax^2+32x+c=0$ has exactly one solution. If a+c=130, and a>c, find the ordered pair (a,c).

Thanks!

 Jun 9, 2019
 #1
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If we have only one solution, the discriminant must  = 0

 

So

 

32^2  - 4ac  = 0             ⇒  c  = 130 - a

 

1024 - 4  (a) (130 - a)  = 0

 

1024 - 520a + 4a^2  = 0

 

4a^2 -520a + 1024  = 0       divide through by 4

 

a^2 - 130a + 256  = 0      factor as

 

(a -128) (a - 2)  = 0

 

Set each factor to 0  and solve for a

 

a - 128  = 0           a - 2 =0

a = 128                  a = 2

 

But...since  a > c

 

Then a must = 128    and c = 130 - 128  = 2  

 

(a, c )  =  (128, 2)

 

cool cool cool

 Jun 9, 2019

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