Question

\[\sqrt{2x^2 +9} - \sqrt{2x^2-9} = \sqrt{a} -\sqrt{b} , and \sqrt{2x^2+9} + \sqrt{2x^2-9} = 9 - 3\sqrt{7}\] then the values of a and b are respectively?

Answer 1

error

Answer 2

where

Answer 3

\[\sqrt{2x^2+9}+\sqrt{2x^2-9}=9-3\sqrt{7}\]\[(\sqrt{2x^2+9}+\sqrt{2x^2-9})(\sqrt{2x^2+9}-\sqrt{2x^2-9})=(9-3\sqrt{7})(\sqrt{2x^2+9}-\sqrt{2x^2-9})\]\[(2x^2+9)-(2x^2-9)=(9-3\sqrt{7})(\sqrt{2x^2+9}-\sqrt{2x^2-9})\]\[18=(9-3\sqrt{7})(\sqrt{2x^2+9}-\sqrt{2x^2-9})\]therefore:

Answer 4

\[\sqrt{2x^2+9}-\sqrt{2x^2-9}=\frac{18}{9-3\sqrt{7}}=\frac{18(9+3\sqrt{7})}{(9-3\sqrt{7})(9+3\sqrt{7}))}\]\[\qquad=\frac{18(9+3\sqrt{7})}{81-73}=\frac{18(9+3\sqrt{7})}{18}=9+3\sqrt{7}\]

Answer 5

so now we have two equations that we can use:\[\sqrt{2x^2+9}+\sqrt{2x^2-9}=9-3\sqrt{7}\]\[\sqrt{2x^2+9}-\sqrt{2x^2-9}=9+3\sqrt{7}\]

Answer 6

the answer is 81, 63 @asnaseer did u got that

Answer 7

if we add these up we get:\[2\sqrt{2x^2+9}=18\]\[\sqrt{2x^2+9}=9\]\[2x^2+9=81\]therefore:\[2x^2-9=63\]therefore:\[\sqrt{2x^2+9}-\sqrt{2x^2-9}=\sqrt{81}-\sqrt{63}\]and you have your answer.

Answer 8

let me check and if i have doubts i will inform u thanzzzzz

Answer 9

yw

Answer 10

i got it now thzz

Answer 11

ok - good :)

Answer 12

I've been checking this result myself as it seemed odd that:
\[\sqrt{2x^2+9}+\sqrt{2x^2-9}=9-3\sqrt{7}\]\[\sqrt{2x^2+9}-\sqrt{2x^2-9}=9+3\sqrt{7}\]
I think your second equation should have been:\[\sqrt{2x^2+9} + \sqrt{2x^2-9} = 9 + 3\sqrt{7}\]

Answer 13

ya it 2 +

Answer 14

u wrote 2*

Answer 15

i.e. the question should have been:
\[\sqrt{2x^2 +9} - \sqrt{2x^2-9} = \sqrt{a} -\sqrt{b} , and \sqrt{2x^2+9} + \sqrt{2x^2-9} = 9 + 3\sqrt{7}\]

Answer 16

no it is 9-3root7..