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Show that x+y >=2 if x is root of ax^2+bx+c=0 and y is root of cy^2+b y +a=0?
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You need to find the solutions to both quadratic equations, using quadratic formula, such that:
`x_(1,2) = (-b +- sqrt(b^2 - 4ac))/(2a)`
`y_(1,2) = (-b +- sqrt(b^2 - 4ac))/(2c)`
You should notice the property of quadratic equations based on its coefficients, such that:
x_(1,2) = 1/(y_(1,2)) => y = 1/x
Hence, replacing `1/x` for `y` in the inequality you need to test, yields:
`x + y = x + 1/x >= 2`
You need to move all terms to one side and then you need to bring the terms to a common denominator, such that:
`x + 1/x - 2 >= 0 => x^2 - 2x + 1 >= 0`
`x^2 - 2x + 1 = (x - 1)^2 >= 0`
Since any square is positive or 0, hence, the inequality `(x - 1)^2 >= 0` holds.
Hence, testing if the inequality `x+y>=2` holds, under the given conditions, yields that the statement `x+y>=2` is valid.
Posted by sciencesolve on June 22, 2013 at 12:31 PM (Answer #1)
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