We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
51
1
avatar+248 

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
avatar+102417 
+3

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

15 Online Users

avatar