Po is trying to solve the following equation by completing the square: $$49x^2+56x-64 = 0.$$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 $a + b + c$?

Thanks

WhichWitchIsWhich
Oct 22, 2017

#5**+1 **

(x - \(\frac47\))^{2} = \(\frac{80}{49}\)

On this one...to get a , b , and c as integers, we can multiply both sides by 7^{2}

7^{2 }* (x - \(\frac47\))^{2} = 7^{2} * \(\frac{80}{49}\)

( 7 * (x - \(\frac47\)) )^{2} = 49 * \(\frac{80}{49}\)

(7x - 4)^{2} = 80

And... 7 + (-4) + 80 = 83

hectictar
Oct 22, 2017

#1**+1 **

Follow the procedure shown at: https://web2.0calc.com/questions/if-we-express-3x-2-6x-2-in-the-form-a-x-h-2-k-what-is-a-h-k

Alan
Oct 22, 2017

#2**+1 **

Thanks really but I'm not sure I understand the answer there... Can you expound please? thanks so much!

WhichWitchIsWhich
Oct 22, 2017

#3**+2 **

49 x^2+ 56x- 64 = 0 divide through by 49

x^2 + 56/49 - 64/49 = 0

x^2 - 8/7 - 64/49 = 0 add 64/49 to both sides

x^2 - 8/7 = 64/49

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

x^2 - 8/7 + 16/49 = 64/49 + 16/49

x^2 - 8/7 + 16/49 = 80 / 49 factor the left side

(x - 4/7)^2 = 80/49

So a = 1, b = -4/7 and c = 80/49

And a + b + c = 1 - 4/7 + 80/49 = [ 49 - 28 + 80] / 49 = 101 / 49

CPhill
Oct 22, 2017

#5**+1 **

Best Answer

(x - \(\frac47\))^{2} = \(\frac{80}{49}\)

On this one...to get a , b , and c as integers, we can multiply both sides by 7^{2}

7^{2 }* (x - \(\frac47\))^{2} = 7^{2} * \(\frac{80}{49}\)

( 7 * (x - \(\frac47\)) )^{2} = 49 * \(\frac{80}{49}\)

(7x - 4)^{2} = 80

And... 7 + (-4) + 80 = 83

hectictar
Oct 22, 2017