What is 4y^2-34y+289=0,x^2=6 ?

The solution to 4y^2-34y+289=0,x^2=6 is

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4y^2-34y+289=0,x^2=6

4 y^{2} - 34 y + 289 = 0, x^{2} = 6

x = - 6, x = - 6, x = 6, x = 6, y = \frac{17}{4} + i \frac{17 3}{4} y = \frac{17}{4} - i \frac{17 3}{4} y = \frac{17}{4} + i \frac{17 3}{4} y = \frac{17}{4} - i \frac{17 3}{4}

Solution steps

[4 y^{2} - 34 y + 289 = 0 x^{2} = 6]

Solve

x^{2} = 6 : x = - 6, x = 6

For

4 y^{2} - 34 y + 289 = 0

, subsitute

x

with

- 6 : y = \frac{17}{4} + i \frac{17 3}{4}, y = \frac{17}{4} - i \frac{17 3}{4}

For

4 y^{2} - 34 y + 289 = 0

, subsitute

x

with

6 : y = \frac{17}{4} + i \frac{17 3}{4}, y = \frac{17}{4} - i \frac{17 3}{4}

Verify solutions by plugging them into the original equations

Therefore, the final solutions for

4 y^{2} - 34 y + 289 = 0, x^{2} = 6

are

x = - 6, x = - 6, x = 6, x = 6, y = \frac{17}{4} + i \frac{17 3}{4} y = \frac{17}{4} - i \frac{17 3}{4} y = \frac{17}{4} + i \frac{17 3}{4} y = \frac{17}{4} - i \frac{17 3}{4}

x^{2} + y^{2} - 12 x + 26 = 0, \frac{( x ^{2} )}{2} - y^{2} = \frac{7}{2}

x + y + x y = - 11, x^{2} + y^{2} + x y = 13

x - y - 4 = 0, x y - 12 = 0

35 x^{2} - 12 x + 1 = 0, 20 y^{2} - 14 y + 2 = 0

y^{2} x - 7 y = 10, x = 6