Given that p is a prime number, rationalise the denominator of 7√p - p^2 / √p^3. Simplify the answer

Question

Given that p is a prime number, rationalise the denominator of 7√p - p^2 / √p^3. Simplify the answer

Answer 1

the problem is somewhat ambiguous.

there maybe some missing or implicit parentheses.

IF this is the problem,

(7sqrp - p^2)/sqr(p^3)

then it

=(7sqrp- p^2)/psqr(p)

divide numerator and denominator by p

=(7/sqrp - p)/sqr(p)

multiply numerator and denominator by sqr(p)

= (7 -(p)sqr(p))/p which has a rational denominator of p

But maybe you meant

7sqr(p) - (p^2/sqr(p^3))

it seems to be written that way, expressly

then

= 7sqr(p) - (p^2/p(sqr(p))

= 7sqr(p) - sqr(p)

= 6sqr(p) = 6sqr(p)/1 with a rational denominator of 1

My guess is that's probably the solution you want.

Or maybe you meant

7sqr(p - p^2)/sqr(p)^3

then

multiply numerator and denominator by sqr(p)

7sqr(p-p^2)sqr(p)/sqr(p)^3(sqr(p))

= 7sqr(p^2-p^3)/p^2 which has a rational denominator

although if p is prime than (p^2-p^3) is negative and the solution is an imaginary number

which suggests this wasn't the problem as intended.

Usually these problems of rationalizing the denominator involve multiplication by a conjugate pair, so maybe there's a missing term in the problem's denominator that somehow got left out in copying or in the original problem?

the point of the p as a prime number means square root of p is an irrational number, and if you want to rationalize the denominator you can't leave sqr(p) in the denominator