Question

Select the two values of x that are roots of this equation.
X2 + 2x - 5 = 0

Answer 1

`x=-1+√(6 )`

`x=-1-√(6 )`

Explanation:

`x=\frac{-b±\sqrt{b^(2)-4ac } }{2a}`

Ignore the A before the ±. It wouldn't let me type it correctly.

x² + 2x - 5 = 0

a = 1

b = 2

c = - 5

`x=\frac{-2±\sqrt{2^(2)-4((1)(-5)) } }{2(1)}`

`x=(-2±√(4-4((1)(-5)) ) )/(2(1))`

`x=(-2±√(4+20 ) )/(2(1))`

`x=(-2±√(24 ) )/(2)`

`x=(-2±√((2)(12) ) )/(2)`

`x=(-2±√((2)(2)(6) ) )/(2)`

`x=(-2±√((2)(2)(2)(3) ) )/(2)`

`x=(-2±(√(2) )(√(2))(√((2)(3) )) )/(2)`

`x=(-2±2√(6 ) )/(2)`

Two separate equations

`x=(-2+2√(6 ) )/(2)`

`x=(-2-2√(6 ) )/(2)`

`x=-1+√(6 )`

`x=-1-√(6 )`