Question

How do you solve `x^ { 2} + 2x - 28= 0`?

Answer 1

`x=-1+sqrt29` or `-1-sqrt29`

Explanation:

In the given equation`x^2+2x-28=0`, the discriminant `b^2-4ac=2^2-4xx1xx(-28)=4+112=116` is not a perfect square. There this cannot be solved by splitting middle term.

Hence we use quadratic formula `x=(-b+-sqrt(b^2-4ac))/(2a)`

Hence `x=(-2+-sqrt116)/2`

= `(-2+-2sqrt29)/2`

= `-1+sqrt29` or `-1-sqrt29`

Answer 2

`x = -1+-sqrt(29)`

Explanation:

This difference of squares identity can be written:

`a^2-b^2 = (a-b)(a+b)`

We will use this later with `a=(x+1)` and `b=sqrt(29)`

Given:

`x^2+2x-28 = 0`

Notice that `x^2+2x` is the first part of the square trinomial:

`x^2+2x+1 = (x+1)^2`

So we can complete the square like this:

`0 = x^2+2x-28`

`color(white)(0) = x^2+2x+1-29`

`color(white)(0) = (x+1)^2-(sqrt(29))^2`

`color(white)(0) = ((x+1)-sqrt(29))((x+1)+sqrt(29))`

`color(white)(0) = (x+1-sqrt(29))(x+1+sqrt(29))`

Hence:

`x = -1+-sqrt(29)`