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Question
If α, β are the zeros of the polynomial f(x) = ax2 + bx + c, then\[\frac{1}{\alpha^2} + \frac{1}{\beta^2} =\]
Options
- \[\frac{b^2 - 2ac}{a^2}\]
- \[\frac{b^2 - 2ac}{c^2}\]
- \[\frac{b^2 + 2ac}{a^2}\]
- \[\frac{b^2 + 2ac}{c^2}\]
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Solution
We have to find the value of `1/alpha^2+1/beta^2`
Given `alpha` and `beta` are the zeros of the quadratic polynomial f(x) = ax2 + bx + c,
`alpha + ß = - (-text{coefficient of x})/(text{coefficient of } x^3)`
`= (-b)/a`
`alphabeta= (\text{Coefficient of x})/(\text{Coefficient of}x^2)`
`= c/a`
We have,
`1/alpha^2+1/beta^2= (1/alpha+1/beta)^2- 2/(alphabeta)`
`1/alpha^2+1/beta^2=(beta/(alphabeta)+alpha/(alphabeta))- 2/(alphabeta)`
`1/alpha^2+1/beta^2=((alpha+beta)/(alphabeta))^2- 2/(alphabeta)`
`1/alpha^2+1/beta^2= (((-6)/a)/(c/a))^2 -2/(c/a)`
`1/alpha^2+1/beta^2= ((-b)/axxa/c)^2- 2/(c/a)`
`1/alpha^2+1/beta^2= ((-b)/cancel(a)xxcancel(a)/c)^2- 2/(c/a)`
`1/alpha^2+1/beta^2=((-b^2)/c)- (2a)/c`
`1/alpha^2+1/beta^2=((-b^2)/c^2)- (2axxc)/(cxxc)`
`1/alpha^2+1/beta^2=((-b^2)/c^2)- (2ac)/(c^2)`
`1/alpha^2+1/beta^2= (b^2 -2ac)/c^2`
Hence, the correct choice is `(b).`
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