practice problems pending

Q1) Solve the cubic equation (9x^3 - 27x^2 + 26x - 8 = 0), given that one of the root of this equation is double the other.

Q2)
If the product of two roots of the equation (4x^4 - 24x^3 + 31x^2 + 6x - 8 = 0) is 1, find all the roots.

Q3 Obtain a polynomial of lowest degree with integral coefficient, whose one of the zeros is sqrt{5} + sqrt{2}.

Solution 1:
Assume a,2a are the roots.
Put the values in the given equation and equate them.
You will get a cubic in 'a' with one root as 0.
a can't be 0 since that would mean -8 = 0
Other roots will be 2/3,13/21
Putting a = 2/3 in the original equation works out nicely and 13/21 doesn't quite fit it.
So roots will be 2/3,4/3,1 = answer.

Solution 2:


Solution 3:
Simplest way to solve such question is to start with
x = given root = sqrt(5) + sqrt(2)
Square both sides:
x^2 = 7 + 2sqrt(10)
x^2 - 7 = 2.sqrt(10)

Again square to eliminate the sqrt on the right side and now you will have a degree 4 polynomial with all integer coefficients and these roots:
sqrt(5) + sqrt(2)
sqrt(5) - sqrt(2)
-sqrt(5) + sqrt(2)
-sqrt(5) - sqrt(2)

So one way to think of it is that all conjugates of a given root are needed to make all coefficients as integers. It's relatively easy to do for sqrt numbers but what if there are cube roots or other surds involved.

Then simply start with x = <given root> and raise the degree as long as any radical is there.
In this case the answer is:
x^4 - 14x^2 + 9 = 0




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