We are given the equation (3x)^2 - 12(3x) + 27 = 0
You can see that after simplifying this, we'll get a quadratic equation, 9x^2 - 26x + 27 = 0. Quite obvious, eh? But we have a quicker and smarter way of solving this, and it is pretty obvious, too.
One can easily notice the significance of (3x) in the equation. If we make a substitution for (3x) in the equation, we'll get a simpler quadratic equation, which would be easier to solve. So, lets assume (3x) = t
=> t^2 - 12t + 27 = 0
This can now be solved by splitting the middle term method.
=> t^2 - 3t - 9t + 27 = 0
=> t(t - 3) - 9(t - 3) = 0
=> (t - 3)(t - 9) = 0
=> (t - 3) = 0; (t - 9) = 0
=> t = 3; t = 9
We had assumed (3x) = t. So, we can say,
=> 3x = 3; 3x = 9
=> x = 1; x = 3
The answer will be x = 1 and x = 3.
Substitutions like this are very common in maths. It might look that the straightforward approach can be quicker than this, but substitutions like this will save a lot of time once you become good at solving Quadratic Equations.
You can see that after simplifying this, we'll get a quadratic equation, 9x^2 - 26x + 27 = 0. Quite obvious, eh? But we have a quicker and smarter way of solving this, and it is pretty obvious, too.
One can easily notice the significance of (3x) in the equation. If we make a substitution for (3x) in the equation, we'll get a simpler quadratic equation, which would be easier to solve. So, lets assume (3x) = t
=> t^2 - 12t + 27 = 0
This can now be solved by splitting the middle term method.
=> t^2 - 3t - 9t + 27 = 0
=> t(t - 3) - 9(t - 3) = 0
=> (t - 3)(t - 9) = 0
=> (t - 3) = 0; (t - 9) = 0
=> t = 3; t = 9
We had assumed (3x) = t. So, we can say,
=> 3x = 3; 3x = 9
=> x = 1; x = 3
The answer will be x = 1 and x = 3.
Substitutions like this are very common in maths. It might look that the straightforward approach can be quicker than this, but substitutions like this will save a lot of time once you become good at solving Quadratic Equations.
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