Question

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

Please include an explanation as well :)​

Answer 1

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.)

Answer 2

${ \underline{ \underline{ \tt{ \green{ \huge{QUESTION}}}}}}$

${ \sf{let \: r \: , \: s \: , \: and \: t \: be \: the \: solutions \: to \: the \: }} \\ { \sf{equation{ \red{ \: \: \: \: \: \: 2 {x}^{3} - 3 {x}^{2} + 4x - 1 = 0}} }} \\ { \sf{find \: the \: { \underline{ \green{ \: \: {r}^{2} + {s}^{2} + {t}^{2} }}}}}$

${ \underline{ \underline{ \tt{ \green{ \huge{SOLUTION}}}}}}$

${ \dashrightarrow{ \underline{ \underline{ \pink{ \sf{Vieta's \: formula}}}}}} \\ \\ { \blue{ \sf{relate \: the \: coefficients \: of \: a \: polynomial }}} \\ { \sf{ \blue{to \: sum \: and \: products \: of \: its \: roots.}}}$

${ \tt{ \underline{given \: equation}{ \green { \: \: \: \: 2 {x}^{3} - 3 {x}^{2} + 4 - 1 = 0}}}} \\ \\ { \tt{ \underline{ comparable \: eqn}{ \green{ \: \: a {x}^{3} + b {x}^{2} + cx + d = 0}}}} \\ \\ { \sf{ \blue{on \: comparing \: both \: . \: we \: get}}}$

${ \dashrightarrow{ \sf{a = { \red{2}}}}} \\ { \dashrightarrow{ \sf{b = { \red{ - 3}}}}} \\ { \dashrightarrow{ \sf{c = { \red{4}}}}} \\ { \dashrightarrow{ \sf{d = { \red{ - 1}}}}}$

${ \underline{ \sf{according \: to \: question}}} \\ \\ { \sf{ \green{the \: roots \: are \: r, \: s, \: and \: t}}}$

${ :{ \implies{ \tt{ \pink{r + s + t = \frac{ - b}{a} = \frac{ - ( - 3)}{2} = \frac{3}{2}}}}}} \\ \\ \\ { : { \implies{ \tt{ \pink{rs + st + rt = \frac{c}{a} = \frac{4}{2} = 2}}}}} \\ \\ \\ { : { \implies{ \tt{ \pink{rst = \frac{ - d}{a} = \frac{ - ( - 1)}{2} = \frac{1}{2} }}}}}$

${ \underline{ \green{ \sf{ {(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca)}}}} \\ \\ { \sf{similarly}} \\ \\ { \underline{ \sf{ \red {{(r + s + t)}^{2} = {r}^{2} + {s}^{2} + {t}^{2} + 2(rs + st + rt)}}}} \\ \\ { \underline{ \sf{ \red{ {r}^{2} + {s}^{2} + {t}^{2} = {(r + s + t)}^{2} - 2(rs + st + rt)}}}}$

${ : { \implies{ \green{ \sf{ {r}^{2} + {s}^{2} + {t}^{2} = {( \frac{3}{2}) }^{2} - 2(2)}}}}} \\ \\ \\ { : { \implies{ \green{ \sf{ {r}^{2} + {s}^{2} + {t}^{2} = \frac{9}{4} - 4}}}}} \\ \\ \\ { : { \implies{ \green{ \sf{ {r}^{2} + {s}^{2} + {t}^{2} = \frac{9 - 16}{4} }}}}}$

${ : { \implies{ \underline{ \boxed{ \red{ \sf{ {r}^{2} + {s}^{2} + {t}^{2} = \frac{ - 7}{4} }}}}}}}$