If r and s are the two distinct real roots of the quadratic equation x^2 - 6x + 4 = 0, wha… — GMAT

If $r$ and $s$ are the two distinct real roots of the quadratic equation $x^2 - 6x + 4 = 0$ , what is the value of $\frac{r^3 + s^3}{r + s}$ ?

The value of

\frac{r^3 + s^3}{r + s}

24

Using Vieta's formulas for

x^2 - 6x + 4 = 0

, the sum of the roots is

r + s = 6

and the product is

rs = 4

. Factoring

r^3 + s^3

gives

(r + s)(r^2 - rs + s^2)

. Dividing by

(r + s)

leaves

r^2 - rs + s^2

, which can be rewritten as

(r + s)^2 - 3rs

. Substituting the Vieta values yields

6^2 - 3(4) = 36 - 12 = 24

Apply Vieta's formulas to the given quadratic equation

x^2 - 6x + 4 = 0

The sum of the roots is

r + s = 6

, and the product of the roots is

rs = 4

For a quadratic equation

x^2 + bx + c = 0

, Vieta's formulas state that the sum of roots is

-b

and the product of roots is

c

Factor the sum of cubes expression

r^3 + s^3

\frac{r^3 + s^3}{r + s} = \frac{(r + s)(r^2 - rs + s^2)}{r + s} = r^2 - rs + s^2

The sum of cubes factors algebraically into

(r + s)(r^2 - rs + s^2)

, and

r + s = 6 \neq 0

allows cancellation.

Express

r^2 - rs + s^2

in terms of

(r + s)

and

rs

r^2 - rs + s^2 = (r + s)^2 - 3rs

Since

(r + s)^2 = r^2 + 2rs + s^2

, subtracting

3rs

yields

r^2 - rs + s^2

Substitute the known values

r + s = 6

and

rs = 4

into the expression.

6^2 - 3(4) = 36 - 12 = 24

Evaluating the algebraic expression yields the final value.

Polynomial Factoring and Vieta's Formulas for Quadratic Equations

Estimated Time:1m 30s

Question