A nice Math Olympiad Algebra Equation | Find the Value of x ?
Solve algebra equation #algebra#olympiad
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Пікірлер: 10
The fastest way of solving this type of equations often seen on KZread videos is adding -6x+9 to both hands. ((x-3)+(3x/(x-3)))^2=25
LHS using (a+b)² - 2ab: (x+3x/(x-3))² - 6*x²/(x-3) - 16 = 0. Equalize denumerator in first parantheses: (x²/x-3)² - 6*x²/(x-3) - 16 = 0 substitute x²/(x-3) = u Now he have u²-6u-16 = 0. u1 = 8 => x1;2 = 4±2i√2. u2 = -2 => x3;4 = -1± √7.
we can also use this method and find easily all solutions in less time.. a=x & b=3x/(x-3)
@argenispipo63
29 күн бұрын
Mucha vuelta inútiles
Write the left hand side of the first equation in form of (a+b)² - 2ab. You will be surprised how quicker it will be to solve then. I will provide detailed solution today if no one else does.
@superacademy247
Ай бұрын
Please do because the Math community needs more methods. Thanks 👍💯 for your support.
You posted the exact same video two months ago: kzread.info/dash/bejne/i22N2NqsmKmcg6Q.html I also posted a comment on your video at the time which presented a much more convenient and concise solution relying on the technique of converting a product of two quantities into a difference of squares and which avoids the need to deal with fractions or substitutions. For convenience sake I'll repost my comment on that video here. Substituting (1) y = 3x/(x − 3) in the original equation gives (2) x² + y² = 16 while from (1) we get (3) 3(x + y) = xy so the real solutions pairs (x, y) are the intersections of a circle and a hyperbola, but I don't see the advantage of an additional trigonometric substitution here. Rewriting (2) as (4) (x + y)² − 2xy = 16 and setting the sum and product of x and y equal to s and p, that is (5) x + y = s (6) xy = p and substituting (5) and (6) in (3) and (4) we have (7) 3s = p (8) s² − 2p = 16 From (7) and (8) we obtain (9) s² − 6s − 16 = 0 which gives s = −2 or s = 8 and so (s, p) = (−2, −6) or (s, p) = (8, 24). Now, (s, p) = (−2, −6) means that x and y are the solutions of the quadratic equation t² + 2t − 6 = 0 which gives x = −1 + √7 ⋁ x = −1 − √7 and (s, p) = (8, 24) means x and y are the solutions of the quadratic equation t² − 8t + 24 = 0 which gives x = 4 + 2i√2 ⋁ x = 4 − 2i√2 A different approach which does not require a substitution and which does not require having to deal with fractions starts 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 x = −1 + √7 ⋁ x = −1 − √7 ⋁ x = 4 + 2i√2 ⋁ x = 4 − 2i√2
@superacademy247
Ай бұрын
I'll create more similar questions and do presentation using your two methods. Thanks 👍💯 for your support and input
@NadiehFan
Ай бұрын
@@superacademy247 Yet another method is to note that (x − 3)² = x² − 6x + 9 and therefore x² = (x − 3)² + 6x − 9 so we can rewrite the equation x² + (3x/(x − 3))² = 16 as (x − 3)² + 6x − 9 + (3x/(x − 3))² = 16 or (x − 3)² + 6x + (3x/(x − 3))² − 25 = 0 Here (x − 3)² + 6x + (3x/(x − 3))² is a perfect square because 6x is twice the product of (x − 3) and 3x/(x − 3) and we also have 25 = 5² so we get ((x − 3) + 3x/(x − 3))² − 5² = 0 which gives us a difference of two squares at the left hand side whereas the right hand side is zero. This allows us to factor the left hand side using the difference of two squares identity and then apply the zero product property. However, before doing so we first multiply both sides by (x − 3)² to eliminate the fraction so we get ((x − 3)² + 3x)² − (5(x − 3))² = 0 (x² − 3x + 9)² − (5x − 15)² = 0 (x² + 2x − 6)(x² − 8x + 24) = 0 x = −1 + √7 ⋁ x = −1 − √7 ⋁ x = 4 + 2i√2 ⋁ x = 4 − 2i√2
Simplify\(x^{2}+\left(\frac{3\cancel{x}}{\cancel{x}}-3 ight)^{2}=16\)Cancel the common factor\(x^{2}+\left(\frac{3\cancel{x}}{\cancel{x}}-3 ight)^{2}=16\)\(x^{2}+(3-3)^{2}=16\)Subtract the numbers\(x^{2}+(\textcolor{#C58AF9}{3-3})^{2}=16\)\(x^{2}+(\textcolor{#C58AF9}{0})^{2}=16\)Evaluate the exponent\(x^{2}+\textcolor{#C58AF9}{0^{2}}=16\)\(x^{2}+\textcolor{#C58AF9}{0}=16\)Add zero\(x^{2}+0=16\)\(x^{2}=16\)Move terms to the left side\(x^{2}=16\)\(x^{2}-16=0\)Replace \(16\) with \(4^{2}\)\(x^{2}-16=0\)\(x^{2}-4^{2}=0\)Use the difference of squares formula\(a^{2}-b^{2}=(a-b)(a+b)\)\(x^{2}-4^{2}=0\)\((x-4)(x+4)=0\)\((x-4)(x+4)=0\) Solution \(x=4\\ x=-4\)