`2x^2+6xy+5y^2=1, then`

Options

A. `|x|le sqrt(5)`
B. `|x|ge sqrt(5)`
C. `y^(2) le 2`
D. `y^(2) le 4`

Answer

1,3

Explanation

` 2x^(2) + 6 xy + 5y^(2) = 1`
Equation (1) can be rewritten as
` 2x^(2) (6y)x + 5y^(2) - 1 = 0 `
Since x is real, we have
` 36y^(2) - 8 (5y^(2) - 1) ge 0 `
or ` y^(2) le 2 `
or ` - sqrt((2) le y le sqrt(2)`
Equation (1) can also be rewritten as
` 5y^(2) +( 6x) y + 2x^(2) - 1 = 0`
Since y is real, we have
` 36 x^(2) - 20(2x^(2) - 1 ) ge 0 `
or ` 36x^(2) - 40 x^(2) + 20 ge 0 `
or ` - 4x^(2) ge - 20 `
or ` x^(2) ke 5 `
or ` - sqrt(5) le x le sqrt(5)` .