Question

How do you solve the quadratic equation by completing the square: `5x^2 + x = −5`?

Answer 1

`x_1 = -1/10 - (3isqrt(11))/10`, `x_2 = -1/10 + (3isqrt(11))/10`

Explanation:

The first thing you need to do is get the equation to the form

`x^2 + b/ax = -c/a`

In your case, `a=5`, so you get

`(color(red)(cancel(color(black)(5))) * x^2)/color(red)(cancel(color(black)(5))) + 1/5x = (-5)/5`

`x^2 + 1/5x = -1`

In order to solve this quadratic by completing the square, you need to write the left side of the equation as the square of a binomial by adding a term to both sides of the equation.

This term will be determined by the coefficient of the `x`-term. What you need to do is divide this coefficient by 2, then square the result.

In your case, you have

`1/5 * 1/2 = 1/10`, then

`(1/10)^2 = 1/100`

The quadratic can thus be written as

`x^2 + 1/5x + 1/100 = -1 + 1/100`

The left side of the equation can now be written as

`x^2 + 1/5x + 1/100 = x^ + 2 *(1/10)x + (1/10)^2 = (x + 1/10)^2`

This means that you have

`(x + 1/10)^2 = -99/100`

Notice that this equation has no real solutions, but it does have two distinct complex solutions.

Take the square root of both sides to get

`sqrt((x + 1/1)^2) = sqrt(-99/100)`

`x + 1/10 = +- sqrt(-99/100) = +- (3sqrt(-11))/10`

Use the fact that the square root of a negative number can be written as

`color(blue)(sqrt(-n) = sqrt((-1) * n) = sqrt((-1)) * sqrt(n) = sqrt(i^2) * sqrt(n) = isqrt(n))`

In your case, the right side of the equaation is equivalent to

`+- (3sqrt((-1) * 11))/10 = +- (3isqrt(11))/10`

This means that you have

`x_(1,2) = -1/10 +- (3isqrt(11))/10`

The two solutions of the quadratic will thus be

`x_1 = color(green)(-1/10 - (3isqrt(11))/10)` and `x_2 = color(green)(-1/10 + (3isqrt(11))/10)`