What is 3x^2-1=4λx,-y=2λx,x^2+y^2=1 ?

The solution to 3x^2-1=4λx,-y=2λx,x^2+y^2=1 is

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3x^2-1=4λx,-y=2λx,x^2+y^2=1

3 x^{2} - 1 = 4 λ x, - y = 2 λ x, x^{2} + y^{2} = 1

x = \frac{1 + 2 7}{3}, x = - \frac{1 + 2 7}{3}, y = - \frac{7 - 1}{3}, y = - \frac{7 - 1}{3}, λ = \frac{7 - 1}{2 1 + 2 7} λ = - \frac{7 - 1}{2 1 + 2 7}

Solution steps

3 x^{2} - 1 = 4 λ x - y = 2 λ x x^{2} + y^{2} = 1

Isolate

y

for

- y = 2 λ x : y = - 2 x λ

Plug

y = - 2 x λ

into

x^{2} + y^{2} = 1

x^{2} + (- 2 x λ)^{2} = 1

Simplify

x^{2} + 4 x^{2} λ^{2} = 1

Solve the system of equations:

3 x^{2} - 1 = 4 λ x, x^{2} + 4 x^{2} λ^{2} = 1 : x = \frac{1 + 2 7}{3}, x = - \frac{1 + 2 7}{3}, x = \frac{1 - 2 7}{9}, x = - \frac{1 - 2 7}{9}, λ = \frac{7 - 1}{2 1 + 2 7} λ = - \frac{7 - 1}{2 1 + 2 7} λ = i \frac{( 5 + 7 ) - 1 + 2 7}{18} λ = - i \frac{( 5 + 7 ) - 1 + 2 7}{18}

Verify solutions by plugging them into the original equations

Therefore, the final solutions for

3 x^{2} - 1 = 4 λ x, - y = 2 λ x, x^{2} + y^{2} = 1

are

x = \frac{1 + 2 7}{3}, x = - \frac{1 + 2 7}{3}, y = - \frac{7 - 1}{3}, y = - \frac{7 - 1}{3}, λ = \frac{7 - 1}{2 1 + 2 7} λ = - \frac{7 - 1}{2 1 + 2 7}

2 a x^{2} + (3 a - 1) x + a + b = 0

x^{2} + 2 x + 7 = x + 7

- x^{2} + 2 x + 9 = 0

d^{2} = 50 - 23 d

0 = - \frac{1}{5} x^{2} - 10

What is 3x^2-1=4λx,-y=2λx,x^2+y^2=1 ?
The solution to 3x^2-1=4λx,-y=2λx,x^2+y^2=1 is