Let P(x) = x^2 - mx + n be a quadratic polynomial with real coefficients m and n, having t… — GRE General Test

Let $P(x) = x^2 - mx + n$ be a quadratic polynomial with real coefficients $m$ and $n$ , having two distinct real roots $\alpha$ and \beta. If the roots satisfy the system of equations $\alpha^3 + \beta^3 = m(n + 7)$ and $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{10}{n^2}$ , which of the following statements MUST be true? Select all such statements.

The product of the roots, $n$ , is equal to $1.5$ .Cevap
The sum of the squares of the roots, $\alpha^2 + \beta^2$ , is equal to $10$ .Cevap
The discriminant of the polynomial $P(x)$ is equal to $7$ .Cevap
D
The absolute value of the sum of the roots, $|\alpha + \beta|$ , is equal to $5$ .
E
The roots $\alpha$ and $\beta$ have opposite signs.

Cevap

The correct statements are those asserting that the product of the roots is

1.5

, the sum of the squares of the roots is

10

, and the discriminant of

P(x)

7

Using Vieta's formulas and algebraic identity expansions for

\alpha^3 + \beta^3

and

\frac{1}{\alpha^2} + \frac{1}{\beta^2}

establishes the system of equations

m^2 = 4n + 7

and

m^2 = 2n + 10

. Solving this system gives

n = 1.5

m^2 = 13

, and a discriminant

\Delta = m^2 - 4n = 7

. Thus, the product of roots is

1.5

, the sum of squares

\alpha^2 + \beta^2 = m^2 - 2n = 10

, and the discriminant is

7

Adım Adım Çözüm

Apply Vieta's formulas to express sum and product of roots.

\alpha + \beta = m \quad \text{and} \quad \alpha\beta = n

Vieta's relations link polynomial coefficients directly to symmetrical root expressions.

Expand

\alpha^3 + \beta^3

in terms of

m

and

n

\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = m^3 - 3mn = m(m^2 - 3n)

Using algebraic identities converts root powers into functions of

m

and

n

Equate the expression from Step 2 to

m(n + 7)

to find an equation for

m^2

m(m^2 - 3n) = m(n + 7) \implies m^2 - 3n = n + 7 \implies m^2 = 4n + 7

Since the roots are distinct,

m \neq 0

, allowing division by

m

Simplify the second given equation

\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{10}{n^2}

\frac{\alpha^2 + \beta^2}{\alpha^2\beta^2} = \frac{m^2 - 2n}{n^2} = \frac{10}{n^2} \implies m^2 - 2n = 10 \implies m^2 = 2n + 10

Combining fractions over a common denominator

(\alpha\beta)^2 = n^2

isolates

m^2 - 2n

Solve for

n

m^2

, and the discriminant

\Delta

4n + 7 = 2n + 10 \implies 2n = 3 \implies n = 1.5; \quad m^2 = 13; \quad \Delta = m^2 - 4n = 13 - 6 = 7

Equating the two expressions for

m^2

yields unique values for

n

m^2

, and

\Delta

Anahtar Kavram

Quadratic Equations, Vieta's Formulas, and Symmetric Polynomial Expressions

Soru

Cevap

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Anahtar Kavram

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