Express √(98-40√6) in the form p√2 + q√3 where p and q are integers.?

Question

Maths problem. Help to solve this please?

Answer 1

98-40√6 = 2p^2+2pq√6 + 3q^2
2p^2+3q^2 = 98
2pq = -40
q = -20/p
2p^2 +1200/p^2 = 98
p^4 - 49p^2 + 600 = 0
(p^2 - 25)(p^2 - 24) = 0
p = 5
q = -4

√(98-40√6) = 5√2 -4√3

Answer 2

Write it as an equation:
√(98-40√6) = p√2 + q√3

Square both sides of the equation:
(98-40√6) = (p√2 + q√3 )^2
(I'll leave it to you to multiply out the terms of the right side of the equation. It's basic algebra using "FOIL" (if you've been taught that acronym).)

Once you multiply them out, you'll have some terms on the right side with no radical, and some that are multiplied by √6. Set the ones that are multiplied by √6 equal to 40 (which is the coefficient of √6 on the left side). And set the terms that are not multiplied by √6 equal to 98 (which is the non-radical term on the left side).

Now you have 2 equations in 2 unknowns, so you should be able to solve for p and q using basic algebra.

Answer 3

Let x = the first expression.

Then
x * x = 98 - 40 / 6

Then 2p 2 + 3 q 2 + 2pq /6

Equating the likes:
98 = 2p 2 + 3 q 2
and
pq = -20

using guess and check ,

p = -4, q = 5
p = 4, q = -5