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NMTC Inter Level Problem 13

$$a,b,c$$ are reals such that $$a-7b+8c=4$$ and $$8a+4b-c=7$$. The value of $$a^2-b^2+c^2$$ is

Options:

(A) $$0$$

(B) $$12$$

(C) $$8$$

(D) $$1$$

Note by Nanayaranaraknas Vahdam
2 years, 9 months ago

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$$a+8c=4+7b$$ (rewriting equation 1)

$$8a-c=7-4b$$ (rewriting equation 2)

$$\therefore (a^2+64c^2+16ac)+(64a^2-16ac+c^2)= (16+56b+49b^2)+(49-56b+16b^2)$$

$$\therefore 65a^2+65c^2=65b^2+65$$

$$\therefore a^2-b^2+c^2=\boxed{1}$$ · 2 years, 9 months ago

here, it is clear that while a,b,c may vary, $$a^2-b^2+c^2$$ remains constant. So,take a=0. we get two variables and two solutions so find the corresponding value of b and c. now substitute in $$a^2-b^2+c^2$$ to get answer. · 2 years, 4 months ago

Rewriting equation 1,

a = 4 - 8c + 7b

hence, 8a = 32 - 64c + 56b

substituitng this value in equation 2,

32 - 64c + 56b + 4b -c = 7

25 = 65c - 60b

5 = 13c - 12b

5 + 5(b-c) = 8c - 7b

substituting this value in equation 1,

a + 5(b-c+1) = 4

now, we can assume that b = c, therefore getting a = -1

Therefore, $${(-1)}^{2} - {b}^{2} + {b}^{2} = \boxed{1}$$ · 2 years, 8 months ago

Although my method is not the correct one, but it is fine ;) :P · 2 years, 8 months ago