Question

74. If alpha and beta are the roots of x² + 4x + 6
= 0 , then value of alpha³+ beta³=​

Answer 1

Step-by-step explanation:

hope this helps you better

Answer 2

Given :

• Equation - $\tt\:x^{2}+4x+6=0$

• $\tt\:\alpha$ and $\tt\:\beta$ are the roots of the given equation.

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To :

• Find the value of $\tt\:\alpha^{3} + \beta^{3}$ .

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Concept to be used :

We would calculate the sum of the roots and then the product of the roots by using formulae . Then after using cubic identity and substituting the values , we will get the value of alpha³+ beta³ .

The formulae to be used are as follows :

❒ $\bf\:Sum~of~the~roots~=~\frac{-b}{a}$

❒ $\bf\:Product~of~the~roots~=~\frac{c}{a}$

❒ $\bf\:a^{3}+b^{3}=(a+b) ^{3}-3ab(a+b)$

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Solution :

Equation given - $\tt\:x^{2}+4x+6=0$

We know general form of equation is $\tt\:ax^{2}+bx+c=0$ .

So comparing the general form of equation and the given equation we get :

• a = 1 , b = +4 , and c = +6

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Calculating the sum of the roots :

$\bf\:Sum~of~the~roots~=~\frac{-b}{a}$

Substituting the values of b and a

$\bf\implies\:Sum~of~the~roots~=~\frac{-4}{1}$

$\small{\mathfrak{\implies\:Sum~of~the~roots~=~-4}}$

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Calculating the product of the roots :

$\bf\:Product~of~the~roots~=~\frac{c}{a}$

Substituting the values of c and a

$\bf\implies\:Product~of~the~roots~=~\frac{6}{1}$

$\small{\mathfrak{\implies\:Product~of~the~roots~=~6}}$

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Now value of alpha³+ beta³ -

$\bf\:\alpha^{3}+\beta^{3}=(\alpha+\beta)^{3} -3 \times \alpha \times \beta (\alpha + \beta)$

Here as we know that $\tt\:\alpha$ and $\tt\:\beta$ are the roots of the given equation.

Therefore $\tt\:\alpha+\beta$ denotes Sum of the roots and $\tt\:\alpha \times \beta$ denotes Products of the roots .

Plugging the values in the identity :

$\sf\implies\:\alpha^{3}+\beta^{3}=(-4)^{3} -3 \times 6 (-4)$

$\sf\implies\:\alpha^{3}+\beta^{3}=-64 -3 \times (-24)$

$\sf\implies\:\alpha^{3}+\beta^{3}=-64 -(-72)$

$\sf\implies\:\alpha^{3}+\beta^{3}=-64 + 72$

$\small{\underline{\boxed{\mathrm{\implies\:\alpha^{3}+\beta^{3}= 8}}}}$ ★

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