Solve for x: f(g(x))=30 | Factor & Simplify: (x+5)^2+6(x+5)(x-4)

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To solve for x in the equation f(g(x))=30, where f(x)=x^2-3x and g(x)=√(x+6), a substitution u=√(x+6) simplifies the problem to the quadratic equation u^2 - 3u - 30 = 0. This can be solved using the quadratic formula to find u, which is then substituted back to find x. For the expression (x+5)^2 + 6(x+5)(x-4), factoring out (x+5) leads to the simplified form (x+5)(x+5 + 6(x-4)). Further simplification can be done from this point. The discussion focuses on solving quadratic equations and factoring expressions effectively.
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If f(x)=x^2-3x and g(x)=in square root of x+6, solve for x if f(g{x))=30

Also, factor this expression and simplify if necessary: (x+5)^2+6(x+5)(x-4)

Thanks.
 
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What have you tried so far? What are you having trouble with?
 
For the first one, I got to, x+6 - 3(square root of x+6) = 30 lol

I don't know what to do next, I think I'm supposed to get rid of the three, but I'm not sure how.
 
Nikola Tesla said:
For the first one, I got to, x+6 - 3(square root of x+6) = 30 lol

I don't know what to do next, I think I'm supposed to get rid of the three, but I'm not sure how.

You can make a substitution to end up with a quadratic equation,

let u = square root of (x+6)

so you end up with

u2 - 3u = 30

You can then solve for u and thn substitue back to solve for x.
 
Nikola Tesla said:
Also, factor this expression and simplify if necessary: (x+5)^2+6(x+5)(x-4)
For the second one, you may want to factor (x + 5) out.
(x + 5)2 + 6(x + 5)(x - 4) = (x + 5) (x + 5 + 6(x - 4)) = ...
Can you go from here? :)
 
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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