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Question: How do you solve \[3{x^2} + 18x + 15 = 0\] by completing the squares?...

How do you solve 3x2+18x+15=03{x^2} + 18x + 15 = 0 by completing the squares?

Explanation

Solution

Completing the squares refers to transforming the given equation to a form like (x+a)2=x2+2ax+a2{(x + a)^2} = {x^2} + 2ax + {a^2} , as this form contains the squared form of the expression by which we can very easily remove the square root and hence, the two values of the variable can be determined.

Complete Step by Step Solution:
The very first step is to find the resemblance between 3x2+18x+15=03{x^2} + 18x + 15 = 0 and x2+2ax+a2{x^2} + 2ax + {a^2} , as if we are able to do so, we would be able to make this quadratic equation as a polynomial with power 11 , and for the roots for such polynomials can be found by equating them to zero.
First of all, the given equation is not in its simplest form.
3x2+18x+15=03{x^2} + 18x + 15 = 0
We just take common 33from the equation and get the equation below,
3(x2+6x+5)=0\Rightarrow 3\left( {{x^2} + 6x + 5} \right) = 0
x2+6x+5=0\Rightarrow {x^2} + 6x + 5 = 0
To make this in the form of (x+a)2{(x + a)^2} , which in this case would be (x+3)2{(x + 3)^2} , we need 44 more.
Therefore, 44 adding to both sides of the equation, we get
x2+6x+5+4=4\Rightarrow {x^2} + 6x + 5 + 4 = 4
x2+6x+9=4\Rightarrow {x^2} + 6x + 9 = 4
Looking at the LHS of the above equation, one can observe it as the expanded form of (x+3)2{(x + 3)^2}
(x+3)2=22\Rightarrow {(x + 3)^2} = {2^2}
Taking square roots on both sides we get
x+3=±2\Rightarrow x + 3 = \pm {2^{}}
Shifting constants on one side, we get
x=23\Rightarrow x = 2 - 3 , x=23x = - 2 - 3
x=1,5\Rightarrow x = - 1, - 5
Therefore, x=1,5x = - 1, - 5

Note: Completing the squares is a method specially used to solve quadratic equations (highest power of the variable is 22).
In this method, the main focus is to transform the given equation in the form of (x+a)2=x2+2ax+a2=b{(x + a)^2} = {x^2} + 2ax + {a^2} = b .
One should be very cautious while removing the square root from both sides of the equation. Students often make a common mistake of just taking the positive value of the quantity obtained after taking the square root and often miss the negative value. This would return a partial answer to any question.
Roots or critical points are those values for which the equation returns zero value.