If α and β are the roots of $x^2-3x+2=0$, then the equation whose roots are α² and β² is:

α + β = 3 and αβ = 2, so the new roots have sum α² + β² = 9 − 4 = 5 and product $(\alpha \beta {)}^2=4$. The equation is $x^2-5x+4=0$. (Squaring the sum to get 9 produces the first option; keeping the old product or the old sum produces the other two.) The original roots are 1 and 2, and their squares 1 and 4 do satisfy $x^2-5x+4=0$.