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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, 7 months ago

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

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

Square and add, you get

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

\(\therefore 65a^2+65c^2=65b^2+65\)

\(\therefore a^2-b^2+c^2=\boxed{1}\) Aditya Raut · 2 years, 7 months ago

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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. Rutwik Dhongde · 2 years, 2 months ago

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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, \({(-1)}^{2} - {b}^{2} + {b}^{2} = \boxed{1}\) Kartik Sharma · 2 years, 7 months ago

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@Kartik Sharma Although my method is not the correct one, but it is fine ;) :P Kartik Sharma · 2 years, 7 months ago

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@Kartik Sharma See that better and accurate one posted above ;) Aditya Raut · 2 years, 6 months ago

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