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[@NadiehFan]

The equation to solve is
x² + (3x/(x − 3))² = 16
A different approach which does not require a substitution and which does not require having to deal with fractions is to start by multiplying both sides by (x − 3)² to get rid of the fraction. We can then proceed as follows
x²(x − 3)² + 9x² = 16(x − 3)²
x²(x² − 6x + 9) + 9x² = 16(x − 3)²
x²(x² − 6x + 18) = 16(x − 3)²
(x² − 3(x − 3) + 3(x − 3))(x² − 3(x − 3) − 3(x − 3)) = 16(x − 3)²
(x² − 3x + 9)² − 9(x − 3)² = 16(x − 3)²
(x² − 3x + 9)² − 25(x − 3)² = 0
(x² − 3x + 9 + 5(x − 3))(x² − 3x + 9 − 5(x − 3)) = 0
(x² + 2x − 6)(x² − 8x + 24) = 0
x² + 2x − 6 = 0 ⋁ x² − 8x + 24 = 0
(x + 1)² = 7 ⋁ (x − 4)² = −8
x + 1 = √7 ⋁ x + 1 = −√7 ⋁ x − 4 = 2i√2 ⋁ x − 4 = −2i√2
x = −1 + √7 ⋁ x = −1 − √7 ⋁ x = 4 + 2i√2 ⋁ x = 4 − 2i√2
The critical step here is converting the product x²(x² − 6x + 18) into a difference of two squares. It is always possible to convert a product of two quantities into a difference of two squares by taking the average of the two quantities and half the difference between the two quantities. Then, we can get the original quantities back by adding half the difference to their average and by subtracting half the difference from their average. Finally, we can use the difference of two squares identity to turn the product of the sum and the difference of the average and half the difference into a difference of two squares.
If we have two quantities p and q, then their average is ½(p + q) and half their difference is ½(p − q) and then p = ½(p + q) + ½(p − q) and q = ½(p + q) − ½(p − q) so applying the difference of two squares identity (a + b)(a − b) = a² − b² with a = ½(p + q), b = ½(p − q) we have
p·q = (½(p + q) + ½(p − q))·(½(p + q) − ½(p − q)) = (½(p + q))² − (½(p − q))²
Here, we have the product x²(x² − 6x + 18). The average of x² and x² − 6x + 18 is ½(x² + x² − 6x + 18) = ½(2x² − 6x + 18) = x² − 3x + 9 = x² − 3(x − 3) and half the difference between x² and x² − 6x + 18 is ½(x² − (x² − 6x + 18)) = ½(6x − 18) = 3x − 9 = 3(x − 3) so we have
x² = x² − 3(x − 3) + 3(x − 3)
x² − 6x + 18 = x² − 3(x − 3) − 3(x − 3)
which gives
x²(x² − 6x + 18) = (x² − 3(x − 3))² − (3(x − 3))² = (x² − 3x + 9)² − 9(x − 3)²
Bringing over 16(x − 3)² from the right hand side to the left hand side the right hand side becomes zero while at the left hand side we then have
(x² − 3x + 9)² − 25(x − 3)²
which is again a difference of two squares. This can be factored into two quadratics using the difference to two squares identity a² − b² = (a + b)(a − b). Since the right hand side is zero, we can then apply the zero product property A·B = 0 ⟺ A = 0 ⋁ B = 0 to obtain two quadratic equations which are easily solved.
Yet another approach is as follows
x² + (3x/(x − 3))² = 16
(x − 3)² + 6x − 9 + (3x/(x − 3))² = 16
(x − 3)² + 6x + (3x/(x − 3))² − 25 = 0
((x − 3) + 3x/(x − 3))² − 5² = 0
((x − 3)² + 3x)² − (5(x − 3))² = 0
(x² − 3x + 9)² − (5x − 15)² = 0
x² + 2x − 6 = 0 ⋁ x² − 8x + 24 = 0
(x + 1)² = 7 ⋁ (x − 4)² = −8
x + 1 = √7 ⋁ x + 1 = −√7 ⋁ x − 4 = 2i√2 ⋁ x − 4 = −2i√2
x = −1 + √7 ⋁ x = −1 − √7 ⋁ x = 4 + 2i√2 ⋁ x = 4 − 2i√2
Here, the critical step is rewriting x² as (x − 3)² + 6x − 9 and noting that
(x − 3)² + 6x + (3x/(x − 3))²
is a perfect square since 6x = 2·(x − 3)·(3x/(x − 3)) is twice the product of (x − 3) and (3x/(x − 3)). So, with a = (x − 3), b = (3x/(x − 3)) and applying the identity a² + 2·a·b + b² = (a + b)² we have
(x − 3)² + 6x + (3x/(x − 3))² = ((x − 3) + 3x/(x − 3))²
Bringing over the constant 16 from the right hand side to the left hand side this means that we can write the left hand side of the equation as a difference of two squares
((x − 3) + 3x/(x − 3))² − 5²
while the right hand side is now zero. To eliminate the fraction we then multiply both sides of the equation by (x − 3)² which turns the left hand side into
((x − 3)² + 3x)² − (5(x −3))²
which gives
(x² − 3x + 9)² − (5x − 15)²
Now the left hand side is again a difference of two squares which can be factored into two quadratics using the difference to two squares identity a² − b² = (a + b)(a − b). Since the right hand side is zero, we can then apply the zero product property A·B = 0 ⟺ A = 0 ⋁ B = 0 to obtain two quadratic equations which are easily solved.

[@laithysokha5415]

❤❤

[@superacademy247]

Thanks 👍💯

[@richardreiter6861]

You take a complicated problem ensuring that the most basic math students can’t follow the logic. Then you take a bunch of steps to find sqrt(-32) for more sophisticated students
A while back you did the value of a quadratic equation solved for (1/x) and with no discussion showed that to solve for X just put 2c in the denominator
That would have been nice to show that proof

[@superacademy247]

I'll take your humble advice seriously. I'd like to be better in explaining Math to students especially those who need help to improve their Math skills

[@l.w.paradis2108]

I frankly thought this was fine.
Please give an example of how you would do it instead.