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Answer:
[tex]2y^2-7y+6=0[/tex]
Step-by-step explanation:
We are given that [tex]\alpha[/tex] and [tex]\beta[/tex] are the zeroes of the polynomial [tex]6y^2-7y+2[/tex]
[tex]y^2-\frac{7}{6}y+\frac{1}{3}[/tex]
We have to find a quadratic polynomial whose zeroes are [tex]1/\alpha[/tex] and [tex]1/\beta[/tex].
General quadratic equation
[tex]x^2-(sum\;of\;zeroes)x+ product\;of\;zeroes[/tex]
We get
[tex]\alpha+\beta=\frac{7}{6}[/tex]
[tex]\alpha \beta=\frac{1}{3}[/tex]
[tex]\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha \beta}[/tex]
[tex]\frac{1}{\alpha}+\frac{1}{\beta}=\frac{7/6}{1/3}[/tex]
[tex]\frac{1}{\alpha}+\frac{1}{\beta}=\frac{7}{6}\times 3=7/2[/tex]
[tex]\frac{1}{\alpha}\times \frac{1}{\beta}=\frac{1}{\alpha \beta}[/tex]
[tex]\frac{1}{\alpha}\times \frac{1}{\beta}=\frac{1}{1/3}=3[/tex]
Substitute the values
[tex]y^2-(7/2)y+3=0[/tex]
[tex]2y^2-7y+6=0[/tex]
Hence, the quadratic polynomial whose zeroes are [tex]1/\alpha[/tex] and [tex]1/\beta[/tex] is given by
[tex]2y^2-7y+6=0[/tex]