a) Given the equation 3x^2 - 4x + 2 = 0, with roots alpha and beta.

Question

a) Given the equation 3x^2 - 4x + 2 = 0, with roots alpha and beta.

Asked on March 28, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-03-28T04:58:34.019Z

Here's the solution for question 1(i): a) Given the equation $3x^2 - 4x + 2 = 0$, with roots $\alpha$ and $\beta$. For a quadratic equation $ax^2 + bx + c = 0$, the sum of roots is $\alpha + \beta = -\frac{b}{a}$ and the product of roots is $\alpha\beta = \frac{c}{a}$. In this case, $a=3$, $b=-4$, $c=2$. Step 1: Find the sum and product of the roots. $$ \alpha + \beta = -\frac{-4}{3} = \frac{4}{3} $$ $$ \alpha\beta = \frac{2}{3} $$ Step 2: Use the identity for $\alpha^3 + \beta^3$. We know that $\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2)$. Also, $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$. Substitute this into the identity: $$ \alpha^3 + \beta^3 = (\alpha + \beta)[(\alpha + \beta)^2 - 2\alpha\beta - \alpha\beta] $$ $$ \alpha^3 + \beta^3 = (\alpha + \beta)[(\alpha + \beta)^2 - 3\alpha\beta] $$ Step 3: Substitute the values of $\alpha + \beta$ and $\alpha\beta$. $$ \alpha^3 + \beta^3 = \left(\frac{4}{3}\right)\left[\left(\frac{4}{3}\right)^2 - 3\left(\frac{2}{3}\right)\right] $$ $$ \alpha^3 + \beta^3 = \left(\frac{4}{3}\right)\left[\frac{16}{9} - 2\right] $$ $$ \alpha^3 + \beta^3 = \left(\frac{4}{3}\right)\left[\frac{16}{9} - \frac{18}{9}\right] $$ $$ \alpha^3 + \beta^3 = \left(\frac{4}{3}\right)\left[-\frac{2}{9}\right] $$ $$ \alpha^3 + \beta^3 = -\frac{8}{27} $$ This shows that $\boxed{\alpha^3 + \beta^3 = -\frac{8}{27}}$. b) We need to find the equation with integral coefficients whose roots are $\frac{1}{\alpha^2}$ and $\frac{1}{\beta^2}$. Let the new roots be $p = \frac{1}{\alpha^2}$ and $q = \frac{1}{\beta^2}$. The general form of a quadratic equation is $x^2 - (p+q)x + pq = 0$. Step 1: Calculate the sum of the new roots, $p+q$. $$ p+q = \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\beta^2 + \alpha^2}{(\alpha\beta)^2} $$ First, find $\alpha^2 + \beta^2$: $$ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta $$ Using values from part a): $\alpha + \beta = \frac{4}{3}$ and $\alpha\beta = \frac{2}{3}$. $$ \alpha^2 + \beta^2 = \left(\frac{4}{3}\right)^2 - 2\left(\frac{2}{3}\right) = \frac{16}{9} - \frac{4}{3} = \frac{16}{9} - \frac{12}{9} = \frac{4}{9} $$ Now substitute this into the sum of new roots: $$ p+q = \frac{\frac{4}{9}}{\left(\frac{2}{3}\right)^2} = \frac{\frac{4}{9}}{\frac{4}{9}} = 1 $$ Step 2: Calculate the product of the new roots, $pq$. $$ pq = \left(\frac{1}{\alpha^2}\right)\left(\frac{1}{\beta^2}\right) = \frac{1}{(\alpha\beta)^2} $$ $$ pq = \frac{1}{\left(\frac{2}{3}\right)^2} = \frac{1}{\frac{4}{9}} = \frac{9}{4} $$ Step 3: Form the new quadratic equation. $$ x^2 - (p+q)x + pq = 0 $$ $$ x^2 - (1)x + \frac{9}{4} = 0 $$ To obtain integral coefficients, multiply the entire equation by 4: $$ 4x^2 - 4x + 9 = 0 $$ The equation with integral coefficients whose roots are $\frac{1}{\alpha^2}$ and $\frac{1}{\beta^2}$ is $\boxed{4x^2 - 4x + 9 = 0}$.

Accepted Answer · 2026-03-28T04:58:34.019Z

Here's the solution for question 1(i): a) Given the equation $3x^2 - 4x + 2 = 0$, with roots $\alpha$ and $\beta$. For a quadratic equation $ax^2 + bx + c = 0$, the sum of roots is $\alpha + \beta = -\frac{b}{a}$ and the product of roots is $\alpha\beta = \frac{c}{a}$. In this case, $a=3$, $b=-4$, $c=2$. Step 1: Find the sum and product of the roots. $$ \alpha + \beta = -\frac{-4}{3} = \frac{4}{3} $$ $$ \alpha\beta = \frac{2}{3} $$ Step 2: Use the identity for $\alpha^3 + \beta^3$. We know that $\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2)$. Also, $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$. Substitute this into the identity: $$ \alpha^3 + \beta^3 = (\alpha + \beta)[(\alpha + \beta)^2 - 2\alpha\beta - \alpha\beta] $$ $$ \alpha^3 + \beta^3 = (\alpha + \beta)[(\alpha + \beta)^2 - 3\alpha\beta] $$ Step 3: Substitute the values of $\alpha + \beta$ and $\alpha\beta$. $$ \alpha^3 + \beta^3 = \left(\frac{4}{3}\right)\left[\left(\frac{4}{3}\right)^2 - 3\left(\frac{2}{3}\right)\right] $$ $$ \alpha^3 + \beta^3 = \left(\frac{4}{3}\right)\left[\frac{16}{9} - 2\right] $$ $$ \alpha^3 + \beta^3 = \left(\frac{4}{3}\right)\left[\frac{16}{9} - \frac{18}{9}\right] $$ $$ \alpha^3 + \beta^3 = \left(\frac{4}{3}\right)\left[-\frac{2}{9}\right] $$ $$ \alpha^3 + \beta^3 = -\frac{8}{27} $$ This shows that $\boxed{\alpha^3 + \beta^3 = -\frac{8}{27}}$. b) We need to find the equation with integral coefficients whose roots are $\frac{1}{\alpha^2}$ and $\frac{1}{\beta^2}$. Let the new roots be $p = \frac{1}{\alpha^2}$ and $q = \frac{1}{\beta^2}$. The general form of a quadratic equation is $x^2 - (p+q)x + pq = 0$. Step 1: Calculate the sum of the new roots, $p+q$. $$ p+q = \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\beta^2 + \alpha^2}{(\alpha\beta)^2} $$ First, find $\alpha^2 + \beta^2$: $$ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta $$ Using values from part a): $\alpha + \beta = \frac{4}{3}$ and $\alpha\beta = \frac{2}{3}$. $$ \alpha^2 + \beta^2 = \left(\frac{4}{3}\right)^2 - 2\left(\frac{2}{3}\right) = \frac{16}{9} - \frac{4}{3} = \frac{16}{9} - \frac{12}{9} = \frac{4}{9} $$ Now substitute this into the sum of new roots: $$ p+q = \frac{\frac{4}{9}}{\left(\frac{2}{3}\right)^2} = \frac{\frac{4}{9}}{\frac{4}{9}} = 1 $$ Step 2: Calculate the product of the new roots, $pq$. $$ pq = \left(\frac{1}{\alpha^2}\right)\left(\frac{1}{\beta^2}\right) = \frac{1}{(\alpha\beta)^2} $$ $$ pq = \frac{1}{\left(\frac{2}{3}\right)^2} = \frac{1}{\frac{4}{9}} = \frac{9}{4} $$ Step 3: Form the new quadratic equation. $$ x^2 - (p+q)x + pq = 0 $$ $$ x^2 - (1)x + \frac{9}{4} = 0 $$ To obtain integral coefficients, multiply the entire equation by 4: $$ 4x^2 - 4x + 9 = 0 $$ The equation with integral coefficients whose roots are $\frac{1}{\alpha^2}$ and $\frac{1}{\beta^2}$ is $\boxed{4x^2 - 4x + 9 = 0}$.