+267
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
+33661 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
+267 Thanks really but I'm not sure I understand the answer there... Can you expound please? thanks so much!
+129933 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
+9481 (x - \(\frac47\))2 = \(\frac{80}{49}\)
On this one...to get a , b , and c as integers, we can multiply both sides by 72
72 * (x - \(\frac47\))2 = 72 * \(\frac{80}{49}\)
( 7 * (x - \(\frac47\)) )2 = 49 * \(\frac{80}{49}\)
(7x - 4)2 = 80
And... 7 + (-4) + 80 = 83 
