Question

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?

Answer 1

Answer:

The answer is 91

Step-by-step explanation:

We have the following equality:

$49x^2+56x-64=a^2x^2+2abx+b^2-c$

Then a must satisfy that $49=a^2$. So,

$a=7$ or $a=-7$.

1) If a=7, then, it follows for the first equality that 56=14b. Then b=4. Finally, substituting b=4 in the first equality we obtain that -64=16-c. So, c= 80. We conclude that

$a+b+c=7+4+80=91$.

As the problem states, we only consider values of a greater than zero. Then 91 is the only solution for $a+b+c$.