solve simultaneously. m^4 + n^4 = 9m^2 n^2 + 1 ...... (i) m + n = 4 ........ (ii)


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Question Number 10222 by Tawakalitu ayo mi last updated on 30/Jan/17

$solve simultaneously .$ $m^{4} + n^{4} = {9 m}^{2} n^{2} + 1 . . . . . . (i)$ $m + n = 4 . . . . . . . . (ii)$
Commented by prakash jain last updated on 01/Feb/17

$Hi the font that u use appears too$ $big . Are u using a large font size ?$
	Answered by prakash jain last updated on 30/Jan/17

	$m^{4} + n^{4} = 9 m^{2} n^{2} + 1$ $m^{4} + n^{4} + 2 m^{2} n^{2} = 11 m^{2} n^{2} + 1$ ${(m^{2} + n^{2})}^{2} = 11 m^{2} n^{2} + 1$ ${(m^{2} + n^{2} + 2 m n - 2 m n)}^{2} = 11 m^{2} n^{2} + 1$ $[{(m + n)}^{2} - 2 m n] = 11 m^{2} n^{2} + 1$ ${(16 - 2 m n)}^{2} = 11 m^{2} n^{2} + 1$ $m n = u$ ${(16 - 2 u)}^{2} = 11 u^{2} + 1$ $From above$ $1 . Solve for quadratic u to get 2 values for m n .$ $2 . Given m + n = 4 and a value for m n you$ $can find m - n and hence solve for m and n$
	Commented by Tawakalitu ayo mi last updated on 30/Jan/17

	$God bless you sir .$
	Answered by arge last updated on 03/Feb/17

	$m^{4} - 9 m^{2} n^{2} + n^{4} - 1 = 0$ $m^{2} (m^{2} - 9 n^{2}) + (n^{2} - 1) (n^{2} + 1) = 0$ $m^{2} (m - 3 n) (m + 3 n) + (n^{2} - 1) (n^{2} + 1) = 0 . . . . . . (i)$ $m = 4 - n . . . . . . (i i)$ ${(4 - n)}^{2} (4 - n - 3 n) (4 - n + 3 n) + (n^{2} - 1) (n^{2} + 1) = 0$ $(16 - 8 n + n^{2}) (4 - 4 n) (4 + n) + (n^{2} - 1) (n^{2} + 1) = 0$ $4 (16 - 8 n + n^{2}) (1 - n) (4 + n) + (n^{2} - 1) (n^{2} + 1) = 0$ $(64 - 32 n + 4 n^{2}) (4 + n - 4 n - n^{2}) + (n^{2} - 1) (n^{2} + 1) = 0$ $(64 - 32 n + 4 n^{2}) (4 - 3 n - n^{2}) + (n^{2} - 1) (n^{2} + 1) = 0$ $256 - 192 n - 64 n^{2^{}} - 128 n + 96 n^{2} + 32 n^{3} + 16 n^{2} - 12 n^{3} - 4 n^{4} + n^{4} - 1 = 0$ $3 n^{4} - 20 n^{3} - 48 n^{2} + 320 n - 255 = 0$ $p o r d i v i s i o n s i n t e t i c a :$ $(n - 1) (3 n^{3} - 17 n^{2} - 65 n + 255) = 0$ $n = 1 ∴∴∴ R t a$ $m = 3 ∴∴∴ R t a$
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