Question

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Answer 1

Question

If α and β are the roots of 2x² - x - 2 = 0, then (α⁻³ + β⁻³ + 2α⁻¹β⁻¹) is equal to

Solution

Since α and β are the roots of given equation, we have

α + β = 1/2 and αβ = -1

(Using, sum of zeroes = -b/a and product of zeroes = c/a for a quadratic equation of the form of ax² + bx + c)

Now,

α⁻³ + β⁻³ + 2α⁻¹β⁻¹ = $\sf \frac{1}{\alpha^3}+\frac{1}{\beta^3} + \frac{2}{\alpha\beta}$

$\sf\implies \frac{\alpha^3+\beta^3}{(\alpha\beta)^3} + \frac{2}{\alpha\beta}$

Put the value of αβ as -1

This gives

$\sf\implies \frac{\alpha^3+\beta^3}{(-1)^3} +\frac{2}{-1}\\\\\implies -(\alpha^3+\beta^3)-2$

Now, use (x³ + y³) = (x + y)(x² + y² - xy)

$\sf\implies - (\alpha+\beta)(\alpha^2+\beta^2 - \alpha\beta) - 2$

Now,

α² + β² = (α + β)² - 2αβ

So, we get

$\sf\implies -(\alpha+\beta)((\alpha+\beta)^2-2\alpha\beta - \alpha\beta) - 2\\\\\implies -(\alpha+\beta)((\alpha+\beta)^2-3\alpha\beta)-2$

Now put value of (α + β) and αβ

$\sf \implies -(\frac{1}{2})((\frac{1}{2})^2-3(-1))-2\\\\\implies -(\frac{1}{2})(\frac{1}{4}+3)-2\\\\\implies -(\frac{1}{2})(\frac{1}{4}+\frac{12}{4})-2\\\\\implies - (\frac{1}{2})(\frac{13}{4})-2\\\\\implies \frac{-13}{8} - 2\\\\\implies \frac{-13}{8} - \frac{16}{8}\\\\\implies \frac{-29}{8}$

Hence the value of α⁻³ + β⁻³ + 2α⁻¹β⁻¹ is (-29/8)