Question

$\bf {Hola\: Brainlians !}$
Here I came back with one interesting question. .Plz help me !!
1 )
If $\alpha, \beta$ are the roots of the Quadratic equation $6x^2-6x+1 =0$ then Find the value of
$\dfrac{1}{2} (a+b\alpha +c\alpha^2+d\alpha^3) +\dfrac{1}{2} (a+b\beta+c\beta^2+d\beta^3)$

Answer 1

Answer:

$\underline{\purple{\ddot{\Mathsdude}}}$

☆Answered by Rohith kumar maths dude

▪Here for the given equation ,

•a+beta=1 and alphabeta=1/6

▪Lets enter to question: -

●1/2(a+bà+cà^2+dà^3)+1/2(a+b beta+c beta^2+cbeta^3)

•=1/2(2a+b+c[alpha+beta^2-2 alpha beta+d[(alpha+beta)^3-3alpha beta(alpha+beta}}]

=1/2[2a+b+c(1-1/3)+d(1-1/2)]

From this we get that,

a+b/2+c/3+d/4.

☆Hope it helps u @lasyasolasya.

Answer 2

Given Expression,

$\sf \: {6x}^{2} - 6x + 1 = 0$

Here,

• Sum of Zeros = -(-6)/6 = 1
• Product of Zeros = 1/6

Consider

$\sf \: \dfrac{1}{2} (a+b\alpha +c\alpha^2+d\alpha^3) +\dfrac{1}{2} (a+b\beta+c\beta^2+d\beta^3) \\ \\ = \sf \: \dfrac{a}{2} + \dfrac{a}{2} + \dfrac{ \alpha b}{2} + \dfrac{ \beta b}{2} + \dfrac{ { \alpha }^{2} c}{2} + \dfrac{ { \beta }^{2}c }{2} + \dfrac{ { \alpha }^{3}d }{2} + \dfrac{ { \beta }^{3}d }{2} \\ \\ = \sf \: a + \dfrac{b}{2} ( \alpha + \beta ) + \dfrac{c}{2} ( { \alpha }^{2} + \beta {}^{2} ) + \dfrac{d}{2} ( { \alpha }^{3} + \beta {}^{3} ) - - - - - (1)$

We know that,

$\sf \: ( \alpha + \beta ) {}^{2} = { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta \\ \\ \longrightarrow \sf \: {1}^{2} = \alpha {}^{2} + { \beta }^{2} + 2 \times \dfrac{1}{6} \\ \\ \longrightarrow \sf \: \alpha {}^{2} + { \beta }^{2} = 1 - \dfrac{1}{3} \\ \\ \longrightarrow \underline{ \boxed{\sf \: \alpha {}^{2} + { \beta }^{2} = \dfrac{2}{3} }}$

Also,

$\sf \: \alpha {}^{3} + { \beta }^{3} = ( \alpha + \beta ) {}^{3} - 3 \alpha \beta ( \alpha + \beta ) \\ \\ \longrightarrow \sf \: \sf \: \alpha {}^{3} + { \beta }^{3} = {1}^{3} - 3 \times \dfrac{1}{6} \times 1 \\ \\ \longrightarrow \sf \: \sf \: \alpha {}^{3} + { \beta }^{3} =1 - \dfrac{1}{2} \\ \\ \longrightarrow \sf \underline{ \boxed{ \sf \: \alpha {}^{3} + { \beta }^{3} = \dfrac{1}{2} }}$

Substituting all the values in expression (1),

$\sf\dfrac{1}{2} (a+b\alpha +c\alpha^2+d\alpha^3) +\dfrac{1}{2} (a+b\beta+c\beta^2+d\beta^3) = a + \dfrac{b}{2} \times 1 + \dfrac{c}{2} \times \dfrac{2}{3} + \dfrac{d}{2} \times \dfrac{1}{2} \\ \\ \longrightarrow \boxed{ \boxed{ \sf \dfrac{1}{2} (a+b\alpha +c\alpha^2+d\alpha^3) +\dfrac{1}{2} (a+b\beta+c\beta^2+d\beta^3) = a + \frac{b}{2} + \dfrac{c}{3} + \dfrac{d}{4} }}$