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Question
If all three zeroes of a cubic polynomial x3 + ax2 – bx + c are positive, then at least one of a, b and c is non-negative.
Options
True
False
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Solution
This statement is False.
Explanation:
Let α, β and γ be the three zeroes of cubic polynomial x3 + ax2 – bx + c.
Then, product of zeroes = `(-("Constant term"))/("Coefficient of" x^3)`
`\implies` αβγ = `c/1`
`\implies` αβγ = `-c` ......(i)
Given that, all three zeroes are positive.
So, the product of all three zeroes is also positive.
i.e., αβγ > 0
`\implies` – c > 0 .....[From (i)]
`\implies` c < 0
Now, sum of the zeroes = α + β + γ
= `(-("Coefficient of" x^2))/("Coefficient of" x^3)`
`\implies` α + β + γ = `a/1 = -a`
But α, β and γ all are positive.
Thus, their sum is also positive.
So, α + β + γ > 0
`\implies` – a > 0
`\implies` a < 0
And sum of the product of the zeroes taken two at a time
= `("Coefficient of" x)/("Coefficient of" x^3)`
= `(-b)/1`
`\implies` αβ + βγ + γα = `- b`
∵ αβ + βγ + αγ > 0
`\implies` `-b > 0`
`\implies` b < 0
∴ All the coefficients a, b and c are negative.
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