Positive solution required!
x 2 − 1 0 x − 2 9 1 + x 2 − 1 0 x − 4 5 1 − x 2 − 1 0 x − 6 9 2 = 0
Find the positive solution to the above equation.
The answer is 13.
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2 solutions
Good Job ! Jolly Good Job !
think u meant a-16 for the denominator of the second fraction
Let y = x − 5 ⇒ y 2 = x 2 − 1 0 x + 2 5 .
⇒ x 2 − 1 0 x − 2 9 1 + x 2 − 1 0 x − 4 5 1 − x 2 − 1 0 x − 6 9 2 = 0
y 2 − 5 4 1 + y 2 − 7 0 1 − y 2 − 9 4 2 = 0
( y 2 − 5 4 ) ( y 2 − 7 0 ) ( y 2 − 9 4 ) ( y 2 − 7 0 ) ( y 2 − 9 4 ) + ( y 2 − 5 4 ) ( y 2 − 9 4 ) − 2 ( y 2 − 5 4 ) ( y 2 − 7 0 ) = 0
( y 2 − 5 4 ) ( y 2 − 7 0 ) ( y 2 − 9 4 ) x 2 − 1 6 4 x + 6 5 8 0 + x 2 − 1 4 8 x + 5 0 7 6 − 2 2 x 2 − 2 4 8 x + 7 5 6 0 = 0
( y 2 − 5 4 ) ( y 2 − 7 0 ) ( y 2 − 9 4 ) − 6 4 y 2 + 4 0 9 6 = 0
⇒ − 6 4 y 2 + 4 0 9 6 = 0 ⇒ 6 4 y 2 = 4 0 9 6 ⇒ y 2 = 6 4 ⇒ y = ± 8
⇒ { y = + 8 y = − 8 ⇒ x − 5 = 8 ⇒ x − 5 = − 8 x = 1 3 x = − 3
Sir, Is there a method to know which dummy variable must be substituted in equation? I substituted in such a way that a= x 2 − 1 0 x − 2 9 and ended up in x = 5 ± 3 6
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I used y = x − 5 because it is a perfect square. It eliminates the middle x term.
Let x 2 − 1 0 x − 2 9 = a
Then the given equation becomes ,
a 1 + a − 1 6 1 − a − 4 0 2 = 0
⇒ a ( a − 1 6 ) ( a − 4 0 ) 6 4 0 − 6 4 a = 0
⇒ 6 4 0 − 6 4 a = 0 ⇒ a = 1 0
⇒ x 2 − 1 0 x − 2 9 = 1 0
⇒ x = 1 3 , − 3
So the ans is 1 3