Question

Suppose that the roots of x3 + 3x2 + 4x − 11 = 0 are a, b and c, and that the roots of x3 + rx2 + 5s + t = 0 are a + b, b + c, c + a. Find t.​

Answer 1

$\large\underline{\sf{Solution-}}$

We know that,

$\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: a {x}^{3} + b {x}^{2} + cx + d, \: then$

$\red{\boxed{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{b}{a}}}}$

$\red{\boxed{ \bf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}}}}$

$\red{\boxed{ \bf{ \: \alpha \beta \gamma = - \dfrac{d}{a}}}}$

Now,

Given that

$\rm :\longmapsto\:a,b,c \: are \: zeroes \: of \: {x}^{3} + {3x}^{2} + 4x - 11$

So,

$\rm :\longmapsto\:a + b + c = - \dfrac{3}{1} = - \: 3$

$\rm :\longmapsto\:ab+ bc + ca = \dfrac{4}{1} = \: 4$

$\rm :\longmapsto\:abc = - \dfrac{( - 11)}{1} = \: 11$

Now, again given that,

$\rm :\longmapsto\:a + b,b + c,c + a \: are \: zeroes \: of \: {x}^{3} + {rx}^{2} + sx + t$

So,

$\rm :\longmapsto\:(a + b)(b + c)(c + a) = - \dfrac{t}{1} = - \: t$

$\bf\implies \:t \: = - (a + b)(b + c)(c + a)$

$\rm \: = \: - (ab + ac + b^{2} + bc)(c + a)$

$\rm \:= - (abc + a {c}^{2}+ b^{2}c + b {c}^{2} + {ba}^{2} + {ca}^{2} + {ab}^{2} + abc)$

On adding and Subtracting abc, we get

$\rm \:= - (abc + a {c}^{2}+ b^{2}c + b {c}^{2} + {ba}^{2} + {ca}^{2} + {ab}^{2} + abc + abc - abc)$

Now on rearranging the terms, we get

$\rm \: = - \bigg(({ba}^{2} + abc + {ca}^{2} ) + ( {ab}^{2} + {cb}^{2} + abc) + ( abc + {bc}^{2} + {ac}^{2}) - abc\bigg)$

$\rm \: = \: - \bigg(a(ab + bc + ca) + b(ab + bc + ca) + c(ab + bc + ca) - abc\bigg)$

$\rm \: = \: - \bigg((a + b + c)(ab + bc + ca) - abc\bigg)$

On substituting the values, evaluated above, we get

$\rm \: = \: - \bigg( - 3 \times 4 - 11\bigg)$

$\rm \: = \: - \bigg( - 12 - 11\bigg)$

$\rm \: = \: - \bigg( - 23\bigg)$

$\rm \: = \: 23$

Hence,

$\bf\implies \:t = 23$

Answer 2

Answer:

the answer's 23