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VARDHAN2810:
so that u get p and q
why you substituted x value as q and -2q
i have said u
keep x in the give eq
i didn't understand please write on paper and post in my one more same question
please why we should take x and where
because I'm new to these kind of problems
ok
thanks
hi vardhan is it true
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Solution :-
x = [√(p + 2q) + √(p - 2q) ] / [√(p + 2q) - √(p - 2q)]
rationalise the denominator
x = [√(p + 2q) + √(p - 2q) ]2 / [(p + 2q) - (p - 2q)]
x = [√(p + 2q) + √(p - 2q) ]2 / 4q. 4q
x = 2p + 2√(p2 - 4q2) 2qx - p
= √(p2 - 4q2)
squaring on both sides
4q2x2- 4pqx + p2 = p2 - 4q2 4q[ qx2 - px + q ] = 0qx2 - px + q = 0
x = [√(p + 2q) + √(p - 2q) ] / [√(p + 2q) - √(p - 2q)]
rationalise the denominator
x = [√(p + 2q) + √(p - 2q) ]2 / [(p + 2q) - (p - 2q)]
x = [√(p + 2q) + √(p - 2q) ]2 / 4q. 4q
x = 2p + 2√(p2 - 4q2) 2qx - p
= √(p2 - 4q2)
squaring on both sides
4q2x2- 4pqx + p2 = p2 - 4q2 4q[ qx2 - px + q ] = 0qx2 - px + q = 0
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