A nice Math Olympiad | Algebra Problem Solving | Square Root Simplification

Math Olympiad | Algebra Problem Solving | Square Root Simplification
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[@walterwen2975]

Square Root Simplification: x = √[5x + (√6x)]; x = ? x² = 5x + (√6x), (√x)⁴ - 5(√x)² - (√6x) = 0, Let: y = √x (√x)⁴ - 5(√x)² - (√6x) = y⁴ - 5y² - (√6)y = y(y³ - 5y - √6) = 0 y³ - 5y - √6 = [y³ - (√6)³] - (5y - 5√6) = (y - √6)[y² + (√6)y + 6 - 5] y⁴ - 5y² - (√6) = y(y³ - 5y - √6) = y(y - √6)[y² + (√6)y + √6 - 1] = 0 y = 0, y - √6 = 0; y = √6 or y² + (√6)y + 1 = 0, y = (- √6 ± √2)/2; y = √x √x = 0, x = 0; √x = √6, x = 6 or √x = (- √6 ± √2)/2, x = [(- √6 ± √2)/2]² x = (6 + 2 ± 2√12)/4 = (8 ± 4√3)/4 = 2 ± √3 Answer check: x = 0: √[5x + (√6x)] = √(0 + √0) = 0 = x; Confirmed x = 6: √(30 + √36) = √(30 + 6) = √36 = 6 = x; Confirmed x = 2 ± √3: √x = (- √6 ± √2)/2, (√6x) = (√6)(√x) √[5(2 ± √3) + (√6)(- √6 ± √2)/2] = √(7 ± 6√3) ≠ 2 ± √3 = x; Failed Final answer: x = 0 or x = 6

[@tunneloflight]

ok. Question.... when you test the 2 +/- sqrt(3) solutions you did not account for the +/- that results when you cancel the square root of the square. If you pursue the other negative versions, they are indeed roots. Is that right? i.e. for 2 - sqrt(3) = sqrt( 1- - 5 * sqrt(3) MINUS (3 - sqrt(3)), It does confirm that it is a root. ditto for 2 + sqrt(3)

[@superacademy247]

They are NOT roots

[@tunneloflight]

@@superacademy247 Why would you (anyone) pick the positive over the negative of an indeterminate that may be either?

[@tunneloflight]

@@superacademy247 On researching further, this is a very interesting case. There is not universal agreement on the meaning of solutions that result in multi-value conditions. Some argue that they are not solutions as they are not unambiguously true. Others argue that any case that is true proves that they are roots. Others still argue that a caveat is needed specifying that the roots are true when using the negative value of the multi-value function in the RHS for this equation. Others still argue that it is an undefined or open case, and resolution of the question about whether these are roots is ambiguous. In sum - it is an unsettled area of math. Fascinating.

[@tunneloflight]

@@superacademy247In control systems engineering there are cases like this that do arise. Three classes of these are: a) relay race, b) indeterminate solution, c) multi-valued solution that may realize differently at different times. There are others. Finding and eliminating these can be difficult. And best practice is to work hard to avoid known situations that can result in this sort of thing. In essence, they are real world examples of similar conditions showing the truth or falsity is conditional and may be arbitrary. That is something to strenuously be avoided in control systems. So there is a deeper meaning to this quandary. Yet that too does not definitely provide a definitive resolution to the question. In some cases, where having a root or solution occur can result in certain outcomes this can be a very good or very bad thing. As with the question, also an unsolved and likely unsolvable dilemma.

[@SidneiMV]

x² = 5x + √(6x) √x = u => x = u² u⁴ - 5u² - (√6)u = 0 u(u³ - 5u - √6) = 0 u = 0 => *x = 0* u³ - 5u - √6 = 0 u³ - 6u + u - √6 = 0 u(u² - 6) + (u - √6) = 0 u(u + √6)(u - √6) + (u - √6) = 0 (u - √6)(u² + (√6)u + 1) = 0 u = √6 => *x = 6* u² + (√6)u + 1 = 0 u = √x = (-√6 ± √2)/2 < 0 => no real solutions

[@donlowell]

Why don't you use a graphics calculator to check your work? All 4 solutions were correct!!!!

[@superacademy247]

The other two solutions are extraneous due to squaring both sides twice to get rid of the radicals

[@prollysine]

x^4-10x^3+25x^2-6x=0 , x(x^3-10x^2+25x-6)=0 , x=0 , by faktoring , x^3-10x^2+25x-6=0 ,, (x-6)(x^2-4x+1)=0 , x-6=0 , x=6 , x^2-4x+1=0 , /// x^2-4x+1=0 , x= 2+sqrt3 , 2-sqrt3 ///, test -> x= 0 , 6 , 0 OK , x=6 , sqrt(5*6+sqrt(6*6))=sqrt(30+sqrt(36)) , sqrt36= |+/-| 6 , /// -6 not a solu. /// , solu. x= 0 , 6 ,