Function Composition: Solving Quadratic Formulas with Radicals

AI Thread Summary
The discussion revolves around solving the quadratic expression (-3x^2-15x-12)(√5). Participants clarify that to handle the radical, it should be multiplied by the solutions of the quadratic once factored. There is confusion regarding the terminology, as one contributor points out that the task is not about solving but rather factoring or multiplying the polynomial by the square root. The importance of having a complete equation for solving is emphasized, as the current expression lacks that structure. Ultimately, understanding the distinction between factoring and solving is crucial for correctly addressing the problem.
dance_sg
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Homework Statement



(-3x^2-15x-12)(√5)

Homework Equations



Quadratic formula?

The Attempt at a Solution


hmm, i know how to solve the first set of bracets, but what do i do with the radical? how do i multiply it to the polynomial?
 
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all you have to do is multiply it to the two solutions
so say the factored quadratic is (x-4)(x+4), so the solutions are 4 and -4. just multiply the rad5 to those two answers.
 
dance_sg said:

Homework Statement



(-3x^2-15x-12)(√5)

Homework Equations



Quadratic formula?

The Attempt at a Solution


hmm, i know how to solve the first set of bracets, but what do i do with the radical? how do i multiply it to the polynomial?

What are you trying to do here? About all you can do with what you've provided is factor the quadratic or multiply the quadratic by sqrt(5). You are not "solving" the first set of brackets - to solve an equation you first off need an equation, which you don't have.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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