Question

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

Answer 1

See a solution process below:

Explanation:

First, we need to expand the squared term on the right side of the equation using the rule:

#(a + b)^2 = a^2 + 2ab + b^2

Substituting `x` for `a` and `2` for `b` gives:

`y = (x + 2)^2 - x - 5`

`y = x^2 + (2 xx x xx 2) + 2^2 - x - 5`

`y = x^2 + 4x + 4 - x - 5`

We can now group and combine like terms to put the equation in standard form for a quadratic.

`y = x^2 + 4x - x + 4 - 5`

`y = x^2 + 4x - 1x + 4 - 5`

`y = x^2 + (4 - 1)x + (4 - 5)`

`y = x^2 + 3x + (-1)`

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

The quadratic formula states:

For `ax^2 + bx + c = 0`, the values of `x` which are the solutions to the equation are given by:

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

Substituting `1` for `a`; `3` for `b` and `-1` for `c` gives:

`x = (-3 +- sqrt(3^2 - (4 * 1 * -1)))/(2 * 1)`

`x = (-3 +- sqrt(9 - (-4)))/2`

`x = (-3 +- sqrt(9 + 4))/2`

`x = (-3 +- sqrt(13))/2`

Or

`x = -3/2 + sqrt(13)/2` and `x = -3/2 - sqrt(13)/2`