Please help!(Vieta's Formulas) Let , , and be solutions to the equati

1.	Please help!(Vieta's Formulas) Let , , and be solutions to the equation . Find . Please include an explanation as well :)
Answer» ong>Solution :- comparing 2x³ - 3x² + 4X - 1 = 0 with ax³ + bx² + cx + d = 0 we get, a = 2 b = (-3) c = 4 d = (-1) since roots are r, s and t , r + s + t = (-b/a) = -(-3/2) = 3/2 rs + st + rt = c/a = 4/2 = 2 rst = - d/a = -(-1)/2 = (1/2) . we know that, (a + b + c)² = a² + b² + c² + 2(ab + bc + CA) then, → (r + s + t)² = r² + s² + t² + 2(rs + st + rt) → r² + s² + t² = (r + s + t)² - 2(rs + st + rt) PUTTING all values we get, → r² + s² + t² = (3/2)² - 2(2) → r² + s² + t² = (9/4) - 4 → r² + s² + t² = (9 - 16)/4 → r² + s² + t² = (-7/4) (Ans.)

Please help!(Vieta's Formulas) Let , , and be solutions to the equation . Find . Please include an explanation as well :)

Answer»

ong>Solution :-

comparing 2x³ - 3x² + 4X - 1 = 0 with ax³ + bx² + cx + d = 0 we get,

since roots are r, s and t ,

we know that,

then,

→ (r + s + t)² = r² + s² + t² + 2(rs + st + rt)

→ r² + s² + t² = (r + s + t)² - 2(rs + st + rt)

PUTTING all values we get,

→ r² + s² + t² = (3/2)² - 2(2)

→ r² + s² + t² = (9/4) - 4

→ r² + s² + t² = (9 - 16)/4

→ r² + s² + t² = (-7/4) (Ans.)