Solving A Rational Equation in Three Ways

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[@NadiehFan]

If we start by multiplying both sides by (x + 2)² to get rid of the fraction we can proceed as follows:
x²(x + 2)² + 4x² = 5(x + 2)²
x²(x² + 4x + 4) + 4x² = 5(x + 2)²
x²(x² + 4x + 8) = 5(x + 2)²
(x² + 2x + 4 − 2(x + 2))(x² + 2x + 4 + 2(x + 2)) = 5(x + 2)²
(x² + 2x + 4)² − 4(x + 2)² = 5(x + 1)²
(x² + 2x + 4)² − 9(x + 2)² = 0
(x² + 2x + 4 − 3(x + 2))(x² + 2x + 4 + 3(x + 2)) = 0
(x² − x − 2)(x² + 5x + 10) = 0
x² − x − 2 = 0 ∨ x² + 5x + 10 = 0
Now we only need to solve these two quadratic equations using the quadratic formula to obtain all four roots of the original equation.
Note that these two quadratic equations are exactly the quadratic equations obtained in the video, but (1) we did not have to deal with fractions, (2) we did not have to use completion of the square and (3) we did not have to use a substitution.
All that was needed was the difference of two squares identity, first to convert the product x²(x² + 4x + 8) at the left hand side into a difference of two squares (x² + 2x + 4)² − 4(x + 2)², and then, after reducing the right hand side to zero, once again to factor the left hand side (x² + 2x + 4)² − 9(x + 2)² into two quadratics.
You can turn any product of two quantities into a difference of two squares by taking their average and rewriting the quantities as their average plus and minus half their difference. Here we had the product of x² and x² + 4x + 8. Their average is x² + 2x + 4 and their difference is 4x + 8 = 4(x + 2) so we have x² = (x² + 2x + 4) − 2(x + 2) and x² + 4x + 8 = (x² + 2x + 4) + 2(x + 2). Using the difference of two squares identity (a − b)(a + b) = a² − b² we can therefore rewrite the product x²(x² + 4x + 8) as (x² + 2x + 4)² − 4(x + 2)².

[@SyberMath]

You are amazing! 🤩

[@pdarmagn]

That's brilliant 😊

[@misterdubity3073]

Seeing (x) and (x+2) I converted these to (y-1) and (y+1); modified method 1 led to:
y(y^3 - 3y - 18) = 0; y = 0 and y = 3 are solutions --> x values {-1, 2}

[@theearth.8880]

Wow! Beautiful, sir!

[@SyberMath]

Glad you like it!

[@roger7341]

Brute force
x^2(x^2 + 4x + 4) + 4x^2 - 5(x^2 + 4x + 4) = 0
x^4 + 4x^3 + 3x^2 - 20x - 20 = 0
By inspection x = -1 and synthetic division by (x + 1)
x^3 + 3x^2 - 20 = 0
By inspection x = 2 and synthetic division by (x - 2)
x^2 + 5x + 10 = 0
Quadratic solution x = (-5 ± i√15)/2

[@MarcelCox1]

Method 1 actually isn't that hard in this case. By testing the (smaller) factors of 20, you will quickly find that -1 and 2 are solutions and you can reduce the equation to a second degree equation which then gives you the 2 complex solutions. Not as elegant than solution 2, but still not so nasty as expected and easier to find because you just use standard polinomial equation reduction techniques.

[@scottleung9587]

I expanded, then used RRT with long division.

[@pramodbind9031]

Nice

[@SyberMath]

Thanks

[@sendai-shimin]

I solved by 2nd method.

[@rakenzarnsworld2]

x = 2 or -1

[@-basicmaths862]

x isn't equal to zero So devide. byx^2 & substitute x+1=t solving We get x={-1,2}

[@chandrikabiswas5829]

X=2, x=-1, others are imaginary solution

[@giuseppemalaguti435]

Old