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Question 764376: 9x to the fourth power minus 25x to the second power plus 16 is equal to zero
Answer by DrBeeee(684)   (Show Source):
You can put this solution on YOUR website! Given:
(1) 9x^4 - 25x^2 + 16 = 0
Let
(2) y = x^2 in (1) and get
(3) 9y^2 - 25y + 16 = 0 which factors into
(4) (9y - 16)*(y - 1) = 0
FOIL (4) to get
(5) 9y*y - 9y - 16y + 16 = 0 or
(6) 9y^2 - 25y + 16 = 0 which is the same as (3)
Setting each factor of (4) equal to zero yields
(7) 9y - 16 = 0 or
(8) y = 16/9 and
(9) y - 1 = 0 or
(10) y = 1
But from (2) we have
(11) x^2 = 16/9 or
(12) x = +-4/3 and
(13) x^2 = 1 or
(14 x = +-1
Therefore, as expected for a fourth order equation we get four roots
(15) x = {-4/3,-1,1,4/3}
Put each value of x of (15) into (1) to check.
Is (9(-4/3)^4 - 25(-4/3)^2 + 16 = 0)?
Is (9(256/81) - 25(16/9) + 16 = 0)?
Is (256/9 - 400/9 + 16 = 0)?
Is ((-144)/9 + 16 = 0)?
Is (-16 + 16 = 0)?
Is (0 = 0)? Yes
Is (9(-1)^4 - 25(-1)^2 + 16 = 0)?
Is (9 - 25 + 16 = 0)?
Is (-16 + 16 = 0)?
Is (0 = 0)? Yes
Is (9(1)^4 - 25(1)^2 + 16 = 0)?
Is (9 - 25 + 16 = 0)?
Is (-16 + 16 = 0)?
Is (0 = 0)? Yes
Is (9(4/3)^4 - 25(4/3)^2 + 16 = 0)?
Is (9(256/81) - 25(16/9) + 16 = 0)?
Is (256/9 - 400/9 + 16 = 0)?
Is ((-144)/9 + 16 = 0)?
Is (-16 + 16 = 0)?
Is (0 = 0)? Yes
Answer: x = {-4/3,-1,1,4/3}
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