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

1 Answer
Jul 28, 2015

Answer:

#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)#