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show that one zero of 8x²-30x+27 is the square of the other

$T h e g i v e n p o l y n o m i a l i s f (x) = 8 x^{2} - 30 x + 27 = 8 x^{2} - 18 x - 12 x + 27 = 2 x (4 x - 9) - 3 (4 x - 9) = (2 x - 3) (4 x - 9) t h u s, t h e z e r o o f f (x) a r e α = \frac{3}{2} & β = \frac{9}{4} n o w, s q u a r e o f α = α^{2} = {(\frac{3}{2})}^{2} = \frac{9}{4} = β h e n c e, p r o v e d$

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