Question

How do you solve `x^ { 2} + 11x = 7`?

Answer 1

`x = -11/2+-sqrt(149)/2`

Explanation:

Method 1 - Completing the square

First subtract `7` from both sides to get:

`x^2+11x-7 = 0`

The difference of squares identity can be written:

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

We will use this with `a=(2x+11)` and `b=sqrt(149)`

To cut down on arithmetic involving fractions, let us multiply the whole equation by `4` before completing the square:

`0 = 4(x^2+11x-7)`

`color(white)(0) = 4x^2+44x-28`

`color(white)(0) = (2x)^2+2(2x)(11)+11^2-149`

`color(white)(0) = (2x+11)^2-(sqrt(149))^2`

`color(white)(0) = ((2x+11)-sqrt(149))((2x+11)+sqrt(149))`

`color(white)(0) = (2x+11-sqrt(149))(2x+11+sqrt(149))`

Hence:

`2x = -11+-sqrt(149)`

So:

`x = -11/2+-sqrt(149)/2`

`color(white)()`
Method 2 - Quadratic formula

Given:

`x^2+11x=7`

Subtract `7` from both sides to get:

`x^2+11x-7 = 0`

This is in the standard form:

`ax^2+bx+c = 0`

with `a=1`, `b=11` and `c=-7`.

We can find the roots using the quadratic formula:

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

`color(white)(x) = (-11+-sqrt(11^2-4(1)(-7)))/(2*1)`

`color(white)(x) = (-11+-sqrt(121+28))/2`

`color(white)(x) = (-11+-sqrt(149))/2`

`color(white)(x) = -11/2+-sqrt(149)/2`