How do I rationalize the numerator in this expression?

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To rationalize the numerator of the expression (sqrt(h^2 + 5h + 4) - 2) / h, one should multiply by the conjugate of the numerator, which is (sqrt(h^2 + 5h + 4) + 2). This process will help eliminate the square root in the numerator. The correct result should match the answer provided in the back, which is (h + 5) / (sqrt(h^2 + 5h + 4) + 2). Care must be taken to correctly apply the conjugate to avoid errors in the final expression.
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The question states

Rationalize each of the following numerators to obtain and equivalent expression.

(sqroot(h^2+5h+4) -2 ) / h

How do i rationalize the numerator here.

I don't remember

thanks
 
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Expand your fraction with the conjugate of the numerator.
 
thanks a million

i feel dumb
 
actually I am not gettin the answer that matches the back

the ans in the back says

(h+5) / (sqroot(h^2+5h+4) +2)



i get like (h^2+5h -4) / h^2
 
Careful about that conjugate!
You are to multiply your fraction with:
1=\frac{\sqrt{h^{2}+5h+4}+2}{\sqrt{h^{2}+5h+4}+2}
 
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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