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) $12$

(D) $1$

#Algebra

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`2 \times 3`	$2 \times 3$
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Comments

$a+8c=4+7b$ (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 - 6 years, 9 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, ${(-1)}^{2} - {b}^{2} + {b}^{2} = \boxed{1}$

Kartik Sharma - 6 years, 9 months ago

Although my method is not the correct one, but it is fine ;) :P

Kartik Sharma - 6 years, 9 months ago

See that better and accurate one posted above ;)

Aditya Raut - 6 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.

Rutwik Dhongde - 6 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$