How do you solve using the completing the square method #3x^2+10x+13#?

1 Answer
Dec 20, 2017

Answer:

Required Solutions are

#color(blue)(x = (-5/3) +- [sqrt(14)/3]i#

Explanation:

We are given the quadratic expression

#color(red)(f(x)=3x^2+10x+13#

We must use Completing The Square Method to find the solutions.

Let us assume that #color(red)(" " 3x^2+10x+13 = 0)#

We will find the solutions in steps.

#color(green)(Step.1)#

In this step,

we will move the constant term to Right-Hand Side (RHS)

Hence, we get

#color(red)(" " 3x^2+10x = -13)#

#color(green)(Step.2)#

Divide each term by 3 to get the coefficient of the #x^2# term as #1#

#(3/3)x^2+(10/3)x=(-13/3)#

#rArr x^2 + (10/3)x = (-13/3)#

#color(green)(Step.3)#

We will add a value to each side

#rArr x^2 + (10/3)x+ color(red)square = -(13/3) + color(red)square#

In the next step, we need to figure out how-to find the value that goes into the #color(red)(RED)# box

#color(green)(Step.4)#

Divide the coefficient of the #x# term by #2# and square it.

The result of the calculation will replace the #color(red)(RED)# box in the next step.

The calculation is done as follows:

Coefficient of the x-term is #(10/3)#

If we divide by 2, we will get #(10/6)#

When we square this intermediate result, we get

#(10/6)^2 rArr (5/3)^2 rArr (25/9)#

Hence, the value #(25/9)# will replace the #color(red)(RED)# box in the next step.

#color(green)(Step.5)#

#rArr x^2 + (10/3)x+ (25/9) = -(13/3) + (25/9)#

We can write the Left-Hand-Side (LHS) as

#rArr (x+color(red)(5/3))^2 = (-39 + 25)/9#

Note that we get the value #color(red)(5/3)# by dividing the coefficient of the x-term by #2#

#color(green)(Step.6)#

We will now work on

#(x+color(red)(5/3))^2 = (-39 + 25)/9# **

We will take Square Root on both sides to simplify:

Hence, we get

#sqrt[x+(5/3)^2] = +-sqrt[(-39 + 25)/9]#

Square root and Square cancel out.

#:.# on simplification we get,

#x+5/3 = +-sqrt[(-14)/9]#

#rArr x+5/3 = +-sqrt(14*(-1))/sqrt(9)#

Note that, in complex number system,

#i = sqrt(-1)#

#i^2 = i*i= (-1)#

Hence,

#rArr x+5/3 = +-sqrt(14*i^2)/3#

#x = -5/3 +-[sqrt(14)/3]i#

Hence, required solutions are

#color(blue)(x = (-5/3) +- [sqrt(14)/3]i#

Hope this helps.