If
Find the value of
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Answered by
5
_________________
Given,
⇒ a + ( 1/a ) = 17/4.
Now,
⇒ { a - ( 1/a ) }² = { a + ( 1/a ) }² - 4 × a × 1/ a
⇒ { a - ( 1/a ) }² = ( 17/4 )² - 4
⇒ { a - ( 1/a ) }² = ( 289 / 16 ) - 4
⇒ { a - ( 1/a ) }² = ( 289 - 64 ) / 16
⇒ { a - ( 1/a ) }² = 225/16
⇒ { a - ( 1/a ) } = √( 225 / 16 )
∴ a - ( 1/a ) = ± 15/4
Hope it helps !
Given,
⇒ a + ( 1/a ) = 17/4.
Now,
⇒ { a - ( 1/a ) }² = { a + ( 1/a ) }² - 4 × a × 1/ a
⇒ { a - ( 1/a ) }² = ( 17/4 )² - 4
⇒ { a - ( 1/a ) }² = ( 289 / 16 ) - 4
⇒ { a - ( 1/a ) }² = ( 289 - 64 ) / 16
⇒ { a - ( 1/a ) }² = 225/16
⇒ { a - ( 1/a ) } = √( 225 / 16 )
∴ a - ( 1/a ) = ± 15/4
Hope it helps !
Answered by
9
Given:
= > a + 1/a = 17/4.
On Squaring both sides, we get
= > (a + 1/a)^2 = (17/4)^2
= > a^2 + 1/a^2 + 2 * a * 1/a = 289/16
= > a^2 + 1/a^2 + 2 = 289/16
= > a^2 + 1/a^2 = 289/16 - 2
= > a^2 + 1/a^2 = 257/16.
We know that (a - 1/a)^2 = a^2 + 1/a^2 - 2 * a * 1/a
= 257/16 - 2
= 225/16
= > (a - 1/a) = 15/4.
Hope this helps!
= > a + 1/a = 17/4.
On Squaring both sides, we get
= > (a + 1/a)^2 = (17/4)^2
= > a^2 + 1/a^2 + 2 * a * 1/a = 289/16
= > a^2 + 1/a^2 + 2 = 289/16
= > a^2 + 1/a^2 = 289/16 - 2
= > a^2 + 1/a^2 = 257/16.
We know that (a - 1/a)^2 = a^2 + 1/a^2 - 2 * a * 1/a
= 257/16 - 2
= 225/16
= > (a - 1/a) = 15/4.
Hope this helps!
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