solve-simultaneously-m-4-n-4-9m-2-n-2-1-i-m-n-4-ii-

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 \dots \dots (i)

m + n = 4 \dots \dots . . (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 \dots \dots (i)

m = 4 - n \dots \dots (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

Facebook Tweet Pin