The roots of the quadratic $2 x^{2} + 3 x - 1 = 0$ $2x^2+3x-1=0$ are $α$ $alpha$ and $β$ $beta$ . Without calculating the roots, find $α^{3} + β^{3}$ $alpha^3+beta^3$ ?

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The roots of the quadratic $2 x^{2} + 3 x - 1 = 0$ $2x^2+3x-1=0$ are $α$ $alpha$ and $β$ $beta$ . Without calculating the roots, find $α^{3} + β^{3}$ $alpha^3+beta^3$ ?

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Answer 1 · 2017-04-01T18:10:34

$α^{3} + β^{3} = - \frac{45}{8}$ $alpha^3 + beta^3 = -45/8$

Explanation:

Suppose the roots of the general quadratic equation:

$a x^{2} + b x + c = 0$ $ax^2+bx+c = 0$

are $α$ $alpha$ and $β$ $beta$ , then using the root properties we have:

$"sum of roots" \ \ \ \ \ \= alpha+beta = -b/a$
$"product of roots" = alpha beta \ \ \ \ = c /a$

So for the given quadratic with roots $alpha$ and $beta$ :

$2x^2+3x-1 = 0$

we know that:

$alpha+beta = -3/2 \ \ \$ ; and $\ \ \ alpha beta = -1/2$

Consider the binomial expression:

$\ \ \ \ \ (alpha+beta)^3 = alpha^3 + 3alpha^2 beta + 3alpha beta^2 + beta^3$
$:. (alpha+beta)^3 = alpha^3 + beta^3+ 3alpha beta(alpha+beta)$
$:. \ alpha^3 + beta^3 = (alpha+beta)^3 - 3alpha beta(alpha+beta)$

Substituting our values of $alpha+beta$ and $alpha beta$ we get:

$alpha^3 + beta^3 = (-3/2)^3 - 3(-1/2)(-3/2)$
$" " = -27/8 - 9/4$
$" " = -45/8$

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The roots of the quadratic $2 x^{2} + 3 x - 1 = 0$ $2x^2+3x-1=0$ are $α$ $alpha$ and $β$ $beta$ . Without calculating the roots, find $α^{3} + β^{3}$ $alpha^3+beta^3$ ?

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The roots of the quadratic 2x^2+3x-1=02x2+3x−1=02x^2+3x-1=0 are alphaαalpha and betaβbeta. Without calculating the roots, find alpha^3+beta^3α3+β3alpha^3+beta^3?

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Explanation:

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The roots of the quadratic $2 x^{2} + 3 x - 1 = 0$ $2x^2+3x-1=0$ are $α$ $alpha$ and $β$ $beta$ . Without calculating the roots, find $α^{3} + β^{3}$ $alpha^3+beta^3$ ?