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3 years ago

25 (a square) - 4 (b square) + 28bc - 49 (c square)

Mathematics

1 answer:

LUCKY_DIMON [66]3 years ago

8 0

Answer:

(5a - [2b - 7c]) and (5a + [2b + 7c])

Step-by-step explanation:

Factor 25a^2 - 4b^2 + 28bc - 49c^2.

Note that - 4b^2 + 28bc - 49c^2 involves the variables b and c, whereas 25a^2 has only one variable. Thus, try to rewrite - 4b^2 + 28bc - 49c^2 as the square of a binomial:

- 4b^2 + 28bc - 49c^2 = -(4b^2 - 28bc + 49c^2), or

-(2b - 7c)^2.

Thus, the original 25a^2 - 4b^2 + 28bc - 49c^2 looks like:

[5a]^2 - [2b - 7c]^2

Recall that a^2 - b^2 is a special product, the product of (a + b) and (a - b). Applying this pattern to the problem at hand, we conclude:

Thus, [5a]^2 - [2b - 7c]^2 has the factors (5a - [2b - 7c]) and (5a + [2b + 7c])

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Write it in a proportion form.

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Do cross multiplication.

$$\Rightarrow{\frac{15}{4}x =15\times\frac{11}{2}$

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