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What is the product of the irrational roots of the equation (2x-1)(2x-3)(2x-5)(2x-7)=9?
A3/2
B4
C3
D3/4
Show Explanation
Let 2x - 4 = y
=> (y + 3) (y + 1) (y - 1) (y - 3) = 9
=> (y^2 - 1) (y^2 - 9) = 9
=> y^4 - 10y^2 + 9 = 9
=> y^2 (y^2 - 10) = 0
=> y = 0 or± √(10)
=> 2x - 4 =± √(10)
=> x = (1/2) [4± √(10)]
=> product of the irrational roots
= (1/2)^2 * [4 + √(10)] [4 - √(10)]
= (1/4) * 6
= 3/2.
Asked In ::
Let 2x - 4 = y
=> (y + 3) (y + 1) (y - 1) (y - 3) = 9
=> (y^2 - 1) (y^2 - 9) = 9
=> y^4 - 10y^2 + 9 = 9
=> y^2 (y^2 - 10) = 0
=> y = 0 or± √(10)
=> 2x - 4 =± √(10)
=> x = (1/2) [4± √(10)]
=> product of the irrational roots
= (1/2)^2 * [4 + √(10)] [4 - √(10)]
= (1/4) * 6
= 3/2.
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Let 2x - 4 = y
= (y + 3) (y + 1) (y - 1) (y - 3) = 9
= (y^2 - 1) (y^2 - 9) = 9
= y^4 - 10y^2 + 9 = 9
= y^2 (y^2 - 10) = 0
= y = 0 or± √(10)
= 2x - 4 =± √(10)
= x = (1/2) [4± √(10)]
= product of the irrational roots
= (1/2)^2 * [4 + √(10)] [4 - √(10)]
= (1/4) * 6
= 3/2.
Read Full Answer
Report Error
Please Login First Click Here
