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____ [38]
3 years ago
7

Po is trying to solve the following equation by completing the square:$$49x^2+56x-64 = 0.$$He successfully rewrites the above eq

uation in the following form:$$(ax + b)^2 = c,$$where $a$, $b$, and $c$ are integers and $a > 0$. What is the value of $a + b + c$?
Mathematics
1 answer:
aniked [119]3 years ago
8 0

The question is...

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:

3.122

Step-by-step explanation:

49x^2+56x-64 = 0

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

49(x^2+8x/7) = 60

introduce a new constant to form (ax+b)^2 form from the multiplier of x

New constant = ((8/7) ÷ 2)^2

= (8/14)^2

Put the constant into equation plus another equal constant with negative value

49(x^2 + 8x/7 + (8/14)^2 - (8/14)^2) = 60

Reduce the following part

x^2 + 8x/7 + (8/14)^2 = (x + 8/14)^2

Put back into equation:

49((x + 8/14)^2) - 49(8/14)^2 = 60

49((x + 8/14)^2) = 60 + 49(8/14)^2

= 60 + 49(64/196)

= 60 + 16 = 76

(x + 8/14)^2 = 76/49

Equating with (ax + b)^2 = c

a = 1, b = 8/14, c = 76/49

a + b + c = 1 + 8/14 + 76/49 = 3.122

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- but we've just shown that x = p/3

That means that x = y and the triangle is Equilateral.

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