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Question
If α, β, γ are the zeros of the polynomial f(x) = ax3 + bx2 + cx + d, then α2 + β2 + γ2 =
Options
- \[\frac{b^2 - ac}{a^2}\]
- \[\frac{b^2 - 2ac}{a}\]
- \[\frac{b^2 + 2ac}{b^2}\]
- \[\frac{b^2 - 2ac}{a^2}\]
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Solution
We have to find the value of `alpha^2+beta^2+y^2`
Given `alpha,beta,y` be the zeros of the polynomial f(x) = ax3 + bx2 + cx + d,
`alpha + ß + y= - (-text{coefficient of }x^2)/(text{coefficient of } x^3)`
`= (-b)/a`
`alphaß +betay+yalpha= (text{coefficient of x})/(text{coefficient of } x^3)`
`= c/a`
Now
`alpha^2+beta^2+y^2=(alpha+beta+y)^2-2(alphabeta+betay+yalpha)`
`alpha^2+beta^2+y^2=((-6)/a)^2-2(c/a)`
`alpha^2+b^2+y^2= (b^2)/(a^2)-(2c)/a`
`alpha^2+beta^2+y^2=(b^2)/(a^2)- (2cxxa)/(axxa) `
`alpha^2+beta^2+y^2=(b^2)/(a^2)- (2ca)/a^2 `
`alpha^2+beta^2+y^2=(b^2)/(a^2)- (b^2-2ac)/a^2`
The value of `alpha^2+beta^2+y^2=( b^2-2ac)/a^2`
Hence, the correct choice is `(d).`
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