Question 703942: Solve x^1/2-5x^1/4+6=0
Answer by DrBeeee(684)   (Show Source):
You can put this solution on YOUR website! Given;
(1) x^(1/2) - 5*x^(1/4) + 6 = 0
It's not easy to work with square and 4th roots, let's make it easy. OK?
Let y = x^(1/4), then y^2 = x^(1/2).
Now substitute y into (1) and get
(2) y^2 - 5*y + 6 = 0
Equation (2) easily factors into
(3) (y - 3)*(y - 2) = 0 which has the two roots
(4) y = {3,2}
Since y = x^(1/4) we have
(5) x = y^4
Now put the roots of y given by (4) into (5) and get the equivalent roots of x as
(6) x = {3^4,2^4) or
(7) x {81,16}
Let's check these roots of x in (1).
Is (81^(1/2) - 5*81^(1/4) + 6 = 0)?
Is (9 - 5*3 + 6 = 0)?
Is (15 - 15 = 0)?
Is (0 = 0)? Yes
Is (16^(1/2) - 5*16^(1/4) + 6 = 0)?
Is (4 - 5*2 + 6 = 0)?
Is (10 - 10 = 0)?
Is (0 = 0)? Yes
Answer: The two roots of the given equation are x = {81,16}
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