Question

solve
${x}^{2} + 2x - 2 = 0$
​

Answer 1

Answer:

this is the answer

I hope this is the right answer

Answer 2

The roots the equation x² + 2x -2 =0 are  -1+√3 and  -1-√3

Explanation:

Given:

x² + 2x -2 =0

To find:

The zeros of the quadratic equation

Formula:

$\alpha =\frac{-b+\sqrt{b^{2}-4ac} }{2a}$

$\beta =\frac{-b-\sqrt{b^{2}-4ac} }{2a}$

We know that

a = coefficient of x²

b = coefficient of x

c = constant

a = 1

b = 2

c = -2

Substitute the values in the formula

$\alpha =\frac{-b+\sqrt{b^{2}-4ac} }{2a}$

$\alpha =\frac{-2+\sqrt{(2)^{2}-4(1)(-2)} }{2(1)}$

$\alpha =\frac{-2+\sqrt{4-4(-2) } }{2}$

$\alpha =\frac{-2+\sqrt{4-(-8)} }{2}$

$\alpha =\frac{-2+\sqrt{4+8 } }{2}$

$\alpha =\frac{-2+\sqrt{12} }{2}$

$\alpha =\frac{-2+\sqrt{2\times2\times3} }{2}$

$\alpha =\frac{-2+2\sqrt{3} }{2}$

$\alpha =\frac{2(-1+1\sqrt{3}) }{2}$

$\alpha =\frac{2(-1+1\sqrt{3}) }{2}$

α = -1+√3

$\beta =\frac{-b-\sqrt{b^{2}-4ac} }{2a}$

$\beta =\frac{-2-\sqrt{(2)^{2}-4(1)(-2)} }{2(1)}$

$\beta =\frac{-2-\sqrt{4-4(-2) } }{2}$

$\beta =\frac{-2-\sqrt{4-(-8)} }{2}$

$\beta =\frac{-2-\sqrt{4+8 } }{2}$

$\beta =\frac{-2-\sqrt{12} }{2}$

$\beta =\frac{-2-\sqrt{2\times2\times3} }{2}$

$\beta =\frac{-2-2\sqrt{3} }{2}$

$\beta =\frac{2(-1-1\sqrt{3}) }{2}$

β = -1-√3

The roots the equation x² + 2x -2 =0 are  -1+√3 and  -1-√3