If alpha and beta are the roots of the equation 3x2 - 8x − 4 = 0, find the value of alpha2 + beta2.

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:
If $\alpha$ and $\beta$ are the roots of the equation $3x^2 - 8x - 4 = 0$ , find the value of $\alpha^2 + \beta^2$ .

Identify the coefficients of the quadratic equation.
The given quadratic equation is $3x^2 - 8x - 4 = 0$ .
Comparing this to the standard form $ax^2 + bx + c = 0$ , we have:
$a = 3$
$b = -8$
$c = -4$
Find the sum and product of the roots.
For a quadratic equation $ax^2 + bx + c = 0$ , the sum of the roots ( $\alpha + \beta$ ) and the product of the roots ( $\alpha\beta$ ) are given by:
$\alpha + \beta = -\frac{b}{a} = -\frac{-8}{3} = \frac{8}{3}$ $\alpha\beta = \frac{c}{a} = \frac{-4}{3}$
Use the identity for $\alpha^2 + \beta^2$ .
We know that $(\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2$ .
Rearranging this identity to find $\alpha^2 + \beta^2$ :
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$
Substitute the values of the sum and product of roots into the identity.
$\alpha^2 + \beta^2 = \left(\frac{8}{3}\right)^2 - 2\left(\frac{-4}{3}\right)$ $\alpha^2 + \beta^2 = \frac{64}{9} - \left(-\frac{8}{3}\right)$ $\alpha^2 + \beta^2 = \frac{64}{9} + \frac{8}{3}$
Add the fractions to find the final value.
To add the fractions, find a common denominator, which is 9.
$\alpha^2 + \beta^2 = \frac{64}{9} + \frac{8 \times 3}{3 \times 3}$ $\alpha^2 + \beta^2 = \frac{64}{9} + \frac{24}{9}$ $\alpha^2 + \beta^2 = \frac{64 + 24}{9}$ $\alpha^2 + \beta^2 = \frac{88}{9}$
The final answer is $\boxed{\frac{88}{9}}$ .

:
Find the coefficient of $x^{11}$ in the expression of $(x^2 + \frac{1}{x})^{10}$ .
Write the general term of the binomial expansion.
The binomial theorem states that the general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by:
$T_{r+1} = \binom{n}{r} a^{n-r} b^r$
In this expression, $a = x^2$ , $b = \frac{1}{x} = x^{-1}$ , and $n = 10$ .
Substitute these values into the general term formula:
$T_{r+1} = \binom{10}{r} (x^2)^{10-r} (x^{-1})^r$
Simplify the powers of $x$ .
$T_{r+1} = \binom{10}{r} x^{2(10-r)} x^{-r}$ $T_{r+1} = \binom{10}{r} x^{20-2r} x^{-r}$
Combine the terms with $x$ :
$T_{r+1} = \binom{10}{r} x^{20-2r-r}$ $T_{r+1} = \binom{10}{r} x^{20-3r}$
Set the exponent of $x$ equal to 11 to find $r$ .
We are looking for the coefficient of $x^{11}$ , so we set the exponent $20-3r$ equal to 11:
$20 - 3r = 11$ $3r = 20 - 11$ $3r = 9$ $r = 3$
Calculate the binomial coefficient for $r=3$ .
The coefficient of $x^{11}$ is $\binom{10}{r}$ when $r=3$ .
$\text{Coefficient} = \binom{10}{3}$ $\binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10!}{3!7!}$ $\binom{10}{3} = \frac{10 \times 9 \times 8 \times 7!}{3 \times 2 \times 1 \times 7!}$ $\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1}$ $\binom{10}{3} = \frac{720}{6}$ $\binom{10}{3} = 120$
The final answer is $\boxed{120}$ .

If alpha and beta are the roots of the equation 3x2 − 8x − 4 = 0, find the value of alpha2 + beta2.

Step-by-step solution

Identify the coefficients of the quadratic equation.

Find the sum and product of the roots.

Use the identity for $\alpha^2 + \beta^2$ .

Substitute the values of the sum and product of roots into the identity.

Add the fractions to find the final value.

Write the general term of the binomial expansion.

Simplify the powers of $x$ .

Set the exponent of $x$ equal to 11 to find $r$ .

Calculate the binomial coefficient for $r=3$ .

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