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# Help question in Math

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Po is trying to solve the following equation by completing the square49x^2+56x-64 = 0.:He successfully rewrites the above equation in the following form(ax + b)^2 = c, :where a,b , and c are integers and a > 0 . What is the value of a + b + c ?

tertre  Mar 11, 2017

#3
+87334
+5

Oops, tertre...

I meant to write

a = 7 , b = 4 and c  = 80 ....so

a + b + c   =  91

CPhill  Mar 11, 2017
#1
+87334
0

49x^2+56x-64 = 0

49x^2 + 56x  =   64  divide through by  49

x^2 + (56/49)x  =  64/49

x^2 + (8/7)x  = 64/49

Take (1/2) of (8/7)   = 8/14  = 4/7   square it   = 16/49   add to both sides

x^2 + (8/7)x + 16/49 =  64/49 + 16/49

x^2 + (8/7)x + 16/49 = 80/49    multiply through by 49

49x^2  + 56x + 16  = 80      factor the left side

(7x + 4)^2   = 80

a = 7, b = 4 , c = 8

a + b + c  =   19

CPhill  Mar 11, 2017
#2
+2765
0

It says wrong, why?

tertre  Mar 11, 2017
#3
+87334
+5

Oops, tertre...

I meant to write

a = 7 , b = 4 and c  = 80 ....so

a + b + c   =  91

CPhill  Mar 11, 2017
#4
+2765
+5

Thanks so much! I love this website!

tertre  Mar 11, 2017