Question

How do you solve `x^2+x=20`?

Answer 1

`x = 4" "` or `" "x = -5`

Explanation:

Here are a few methods:

`color(white)()`
Method 1 - Fishing for factors

Note that `x^2+x = x(x+1)`, so the given equation can be rewritten:

`x(x+1) = 20`

So we want to find a number `x`, which when multipled by `x+1` gives `20`.

It should not take too much guessing to find:

`color(blue)(4)(color(blue)(4)+1) = (4)(5) = 20`

So `x=4` is a solution.

This is a quadratic equation, so should normally have two solutions, so what is the other?

With a bit of thought, we find:

`((color(blue)(-5))((color(blue)(-5))+1) = (-5)(-4) = 20`

So `x=-5` is the other solution.

`color(white)()`
Method 2 - Completing the square

Given:

`x^2+x=20`

Note that in general:

`(x+p)^2 = x^2+2px+p^2`

So to make `x^2+x` into a perfect square, we need to choose `p=1/2` and add `p^2 = 1/4` to both sides of our equation to get:

`x^2+x+1/4 = 81/4`

Both sides are now perfect squares:

`(x+1/2)^2 = (9/2)^2`

Hence:

`x+1/2 = +-9/2`

Subtracting `1/2` from both sides:

`x = -1/2+-9/2`

That is:

`x = 4" "` or `" "x = -5`

`color(white)()`
Method 3 - Quadratic formula

Given:

`x^2+x=20`

Subtract `20` from both sides to get:

`x^2+x-20 = 0`

This is in the standard form:

`ax^2+bx+c = 0`

with `a=1`, `b=1` and `c=-20`.

It has solutions given by the quadratic formula:

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

`color(white)(x) = (-1+-sqrt((-1)^2-4(1)(-20)))/(2*1)`

`color(white)(x) = (-1+-sqrt(1+80))/2`

`color(white)(x) = (-1+-9)/2`

That is:

`x = 4" "` or `" "x = -5`

Answer 2

`4, - 5`

Explanation:

Method 4. Solving by new Transforming Method (Google Search)
`y = x^2 + x - 20 = 0`
Find 2 real roots, that have opposite signs (ac < 0), knowing the sum (- b = - 1) and the product (c = - 20).
They are: `4 and - 5`