Answer:
4y² - 8y + 3 = 0
Step-by-step explanation:
given 3y² - 8y + 4 ← in standard form
with a = 3, b = - 8, c = 4
and that α and β are roots of the polynomial, then
α + β = - [tex]\frac{b}{a}[/tex] = [tex]\frac{8}{3}[/tex], and
αβ = [tex]\frac{c}{a}[/tex] = [tex]\frac{4}{3}[/tex]
sum of new roots = [tex]\frac{1}{\alpha }[/tex] + [tex]\frac{1}{\beta }[/tex]
= [tex]\frac{\alpha +\beta }{\alpha \beta }[/tex]
= [tex]\frac{\frac{8}{3} }{\frac{4}{3} }[/tex] = 2
product of new roots = [tex]\frac{1}{\alpha }[/tex] × [tex]\frac{1}{\beta }[/tex]
= [tex]\frac{1}{\alpha \beta }[/tex] = [tex]\frac{1}{\frac{4}{3} }[/tex] = [tex]\frac{3}{4}[/tex]
Hence the required equation is
y² - 2y + [tex]\frac{3}{4}[/tex] = 0 ( multiply through by 4 )
4y² - 8y + 3 = 0
