Question

Y = x2 - 2x -3 find the values of x when y = 1

Answer 1

Explanation:

`y=x^2-2x-3`

When
`y=1`,

`x^2-2x-3=1`

`x^2-2x-4=0`

Using quadratic equation,

`x=(-b\pm√(b^2-4ac) )/(2a)`

`x=(-(-2)\pm√((-2)^2-4*1*(-4)) )/(2*1)`

`=(2\pm√(4+16) )/(2)`

`=(2\pm√(20) )/(2)`

`=(2\pm2√(5) )/(2)`

`=1\pm√(5)`

Solution:
`x=1+√(5) ,1-√(5)`