Question

(Vieta's Formulas) Let $\sf r$, $\sf s$, and $\sf t$ be solutions to the equation $2x^3 - 3x^2 + 4x - 1 = 0$. Find $r^2+s^2+t^2$.​

Answer 1

Answer:

Here is the answer

Vieta's Formula for Quadratics:

Vieta's Formula for Quadratics:Given

f(x) = ax 2+bx+c f(x) = ax^2+bx+c

f(x)=a\times2+bx+c , if the equation f(x) = 0 f(x)

= 0 f(x)=0 has roots r 1 r_1 r1 and r 2 r_2

=

r2, then. r 1 + r2=-ba, r1 r2 = ca.

Step-by-step explanation:

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