
How do you solve $ 6{n^2} - 18n - 18 = 6 $ by factoring?
Answer
450k+ views
Hint: In the above question, we are provided with the quadratic equation because the maximum power of variable “n” is two. We have to find the values of “n” by factoring method. So, firstly let two numbers whose addition is coefficient of “n” and multiplication is coefficient of constant term and then by taking common find the roots.
Complete step by step solution:
In this question, we have to find the roots of the quadratic equation with “n” as its variable. The roots are found by the factoring method.
So, the first step is to take the constant at the left-hand side.
$ 6{n^2} - 18n - 18 - 6 = 0 $ or $ 6{n^2} - 18n - 24 = 0 $
We can take common $ 6 $ (this number would not create any change in the solution. So, we can neglect it), then the resultant equation is $ {n^2} - 3n - 4 = 0 $
Let us consider two numbers whose addition is negative of ratio of the coefficient of the $ n $ to the coefficient of the $ {n^2} $ and the multiplication is the ratio of the coefficient of the constant term to the coefficient of the $ {n^2} $
The two such numbers are $ - 4,1 $
Now, splitting the middle terms with these numbers
$ {n^2} - 4n + n - 4 = 0 $
Taking common,
$ n(n - 4) + 1(n - 4) = 0 $
$ (n - 4)(n + 1) = 0 $
Equating to zero, the two brackets
$ n = 4 $ and $ n = - 1 $
These are our required solutions.
So, the correct answer is $ n = 4 $ and $ n = - 1 $ ”.
Note: Be careful about choosing the numbers for factoring. This method is easy but sometimes students get confused for choosing these numbers. Then we can use the formula for finding the roots but if it is mentioned in the question to find the roots by factoring method then use this method only.
Complete step by step solution:
In this question, we have to find the roots of the quadratic equation with “n” as its variable. The roots are found by the factoring method.
So, the first step is to take the constant at the left-hand side.
$ 6{n^2} - 18n - 18 - 6 = 0 $ or $ 6{n^2} - 18n - 24 = 0 $
We can take common $ 6 $ (this number would not create any change in the solution. So, we can neglect it), then the resultant equation is $ {n^2} - 3n - 4 = 0 $
Let us consider two numbers whose addition is negative of ratio of the coefficient of the $ n $ to the coefficient of the $ {n^2} $ and the multiplication is the ratio of the coefficient of the constant term to the coefficient of the $ {n^2} $
The two such numbers are $ - 4,1 $
Now, splitting the middle terms with these numbers
$ {n^2} - 4n + n - 4 = 0 $
Taking common,
$ n(n - 4) + 1(n - 4) = 0 $
$ (n - 4)(n + 1) = 0 $
Equating to zero, the two brackets
$ n = 4 $ and $ n = - 1 $
These are our required solutions.
So, the correct answer is $ n = 4 $ and $ n = - 1 $ ”.
Note: Be careful about choosing the numbers for factoring. This method is easy but sometimes students get confused for choosing these numbers. Then we can use the formula for finding the roots but if it is mentioned in the question to find the roots by factoring method then use this method only.
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