First of all Let x on the LHS brought to RHS. Then the equation becomes 10-x= (28-x√x)^2)^1/3. Now rise it by multiplying thrice then LHS now written as (10-x) ^3= (28-x√x)^2. Expand this. After expanding the term involving root of the function keep either on LHS or on RHS. And again square them. You get a lenear equation .you get the value of X. From that.
As with so many problems of this type, observation quickly identifies major roots. There is a symmetry in the two terms. In this case simply examining for potential cube roots (with a squaring of the second term) suggest 1 and 3^2=9 as the symmetric pair. Checking - they work. The only other potential is 2^2=4 which requires a symmetric pairing with 5. (for both positive pairs). That does not work. Do in the positive pair range, x = 1 and 9 for the only positive case. No solutions exist for the negative real case. That leaves imaginary solutions. Your method or others are required to find those solutions. They too must be symmetric, but in a different way. The imaginary components must balance and negate each other for the two terms.
X+ 3√(28-x√x)²= 10 3√(28-x√x)² = (10 -x)³ Cubing both sides , get (28 - x√x)²= (10-x)³ 784 +x³- 56x√x =1000 - x³-300x + 30x² 2x³ - 30x² +300x -56x√x -216 =0 Clearly from here x=1 is the solution of the equation, after dividing the equations by x-1 , the equation comes 2x²-28x -56√x+272 =0 after applying the quadratic formula the x =9
O mais difícil nesse tipo de questão é encontrar a substituição certa, para o seu desenvolvimento. (The most difficult thing about this type of issue is finding the right replacement for your development.) Congratulations.
First of all Let x on the LHS brought to RHS. Then the equation becomes 10-x= (28-x√x)^2)^1/3. Now rise it by multiplying thrice then LHS now written as (10-x) ^3= (28-x√x)^2. Expand this. After expanding the term involving root of the function keep either on LHS or on RHS. And again square them. You get a lenear equation .you get the value of X. From that.
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As with so many problems of this type, observation quickly identifies major roots. There is a symmetry in the two terms. In this case simply examining for potential cube roots (with a squaring of the second term) suggest 1 and 3^2=9 as the symmetric pair. Checking - they work. The only other potential is 2^2=4 which requires a symmetric pairing with 5. (for both positive pairs). That does not work. Do in the positive pair range, x = 1 and 9 for the only positive case. No solutions exist for the negative real case. That leaves imaginary solutions. Your method or others are required to find those solutions. They too must be symmetric, but in a different way. The imaginary components must balance and negate each other for the two terms.
X+ 3√(28-x√x)²= 10
3√(28-x√x)² = (10 -x)³
Cubing both sides , get
(28 - x√x)²= (10-x)³
784 +x³- 56x√x =1000 - x³-300x + 30x²
2x³ - 30x² +300x -56x√x -216 =0
Clearly from here x=1 is the solution of the equation, after dividing the equations by x-1 , the equation comes 2x²-28x -56√x+272 =0 after applying the quadratic formula the x =9
O mais difícil nesse tipo de questão é encontrar a substituição certa, para o seu desenvolvimento. (The most difficult thing about this type of issue is finding the right replacement for your development.) Congratulations.
It is x > 0.
After
2 x^3 - 30 x^2 + 300 x - 56 x√x - 216 = 0 . (*)
Let √x = t (#).
The (*) rewritten
2 t^6 - 30 t^4 - 56 t^3 + 300 t^2 - 216 = 0 ( because x = t^2 .. ) =>
(t -3)(t-1)(t^4 + 4t^3 -2t^2 - 48t- 36)=0 ( by Horner) => t = 3 , t =1 . .
From (#) => x = 9 and x = 1 .. .
x=1
X= 9; 1
X=9
x+x^3 ➖ ({784➖ x^2}➖ x^2)=x+x^3 ➖ ({782 ➖ x^2})=x+(x^3 )^2➖ 780=x+{x^9 ➖ 780}=771 {x+x ➖ }=x^2+771}= 771x^2 10^70^71x^2 10^7^10^71^1x^2 2^5^7^1^2^5^1^1x^2 1^1^1^1^1^1x^2 1x^2 (x ➖ 2x+1).