Question

If alpha and beta are the roots of 3x²-4x+1. Find

a) alpha²-beta²
b) 2/alpha²+2/beta²
c)2/alpha²-2/beta²​

Answer 1

From Vieta's formulas, we have

`3x^2 - 4x + 1 = 3 (x - \alpha) (x - \beta) \\\\ ~~~~~~~~~~~~~~~~~ = 3x^2 - 3 (\alpha + \beta) + 3\alpha\beta \\\\ \implies \begin{cases} \boxed{\alpha + \beta = \frac43} \\\\ \boxed{\alpha\beta = \frac13} \end{cases}`

a) Note that

`(\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2 \\\\ \implies \alpha^2 - \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta - 2\beta^2`

We have exact values for
`\alpha+\beta` and
`2\alpha\beta`, but sadly not for
`-2\beta^2`. This means we'll need to solve for
`\beta` explicitly.

By factorizing, we have

`3x^2 - 4x + 1 = (3x - 1) (x - 1) = 0 \\\\ \implies x = \frac13 \text{ or } x = 1`

Then the value of
`\alpha^2-\beta^2` depends on which is chosen to be the larger root, and

`\alpha^2 - \beta^2 = \frac1{3^2} - 1^2 = \boxed{-\frac89} \text{ or } \alpha^2 - \beta^2 = 1^2 - \frac1{3^2} = \boxed{\frac89}`

While we're at it, we can also immediately find

`\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(\frac43\right)^2 - 2\cdot\frac13 = \frac{10}9`

b) Combine fractions to get

`\frac2{\alpha^2} + \frac2{\beta^2} = (2\beta^2 + 2\alpha^2)/(\alpha^2\beta^2) \\\\ ~~~~~~~~~~~~~ = 2 \cdot (\alpha^2 + \beta^2)/((\alpha\beta)^2) \\\\ ~~~~~~~~~~~~~ = 2\cdot \frac{\frac{10}9}{\left(\frac13\right)^2} \\\\ ~~~~~~~~~~~~~ = \boxed{20}`

c) Combine fractions again.

`\frac2{\alpha^2} - \frac2{\beta^2} = 2 \cdot (\beta^2 - \alpha^2)/((\alpha\beta)^2) \\\\ ~~~~~~~~~~~~~ = -2\cdot (\pm\frac89)/(\left(\frac13\right)^2) \\\\ ~~~~~~~~~~~~~ = \boxed{\pm16}`