How do you find the roots, real and imaginary, of `y=x^2-5x + 6` using the quadratic formula?

The quadratic formula works for equations of the form:

`y = ax^2 + bx + c`

And states that the solutions for `x` are:

`x = (-b+-sqrt(b^2-4ac))/(2a)`

In this case, the quadratic equation is `y = x^2 - 5x + 6`, so:

`a = 1`
`b = -5`
`c = 6`

So, just plug these values into the quadratic equation and simplify.

`x = (-b+-sqrt(b^2-4ac))/(2a)`

`x = (-(-5)+-sqrt((-5)^2-4(1)(6)))/(2(1))`

`x = (5+-sqrt(25-24))/2`

`x = (5+-sqrt1)/2`

`x = (5-1)/2 color(white)"XX" or color(white)"XX" x = (5+1)/2`

`x = 2 color(white)"XX" or color(white)"XX" x = 3`

Therefore, the solutions to this quadratic are `x=2` and `x=3`. You can write this any way that you like (or whichever way your teacher prefers), but a common way to write it is like this, using set notation:

`x in {2,3}`

This just means that `x` could be any number in the set `{2,3}` to make the equation true.

Final Answer

The standard form of a quadratic function is.

`• y=ax^2+bx+c ; a!=0`

`"For " y=x^2-5x+6`

`rArra=1,b=-5" and " c=6`

Using the `color(blue)"quadratic formula"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(x=(-b+-sqrt(b^2-4ac))/(2a))color(white)(2/2)|)))`

`rArrx=(-(color(red)(-5))+-sqrt(25-(4xx1xx6)))/2`

`color(white)(rArrx)=(5+-sqrt1)/2`

`rArrx_1=(5+1)/2" or " x_2=(5-1)/2`

`rArr" roots are " x_1=3" or " x_2=2`

Both roots are real. There are no imaginary roots.