If the quadratic equation 3x^2 - 12x + c = 0 has two real roots, r_1 and r_2, such that r_… — GRE General Test

If the quadratic equation $3x^2 - 12x + c = 0$ has two real roots, $r_1$ and $r_2$ , such that $r_1^2 + r_2^2 = 10$ , what is the value of the constant $c$ ?

A
3
B
6
9Cevap
D
18
E
-9

Cevap

The value of the constant

c

is 9.

The correct answer is 9. By Vieta's formulas, the sum of the roots is

r_1 + r_2 = -\frac{-12}{3} = 4

and the product of the roots is

r_1 r_2 = \frac{c}{3}

. Using the identity

r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1 r_2

, we substitute the given values to get

10 = 16 - \frac{2c}{3}

, which simplifies to

\frac{2c}{3} = 6

, giving

c = 9

Adım Adım Çözüm

Apply Vieta's formulas to express the sum and product of the roots in terms of equation coefficients.

r_1 + r_2 = -\frac{-12}{3} = 4

and

r_1 r_2 = \frac{c}{3}

For any quadratic equation

ax^2 + bx + c = 0

, the sum of the roots is

-\frac{b}{a}

and the product of the roots is

\frac{c}{a}

Relate the sum of squares

r_1^2 + r_2^2

(r_1 + r_2)

and

r_1 r_2

r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1 r_2

Expanding

(r_1 + r_2)^2 = r_1^2 + 2r_1 r_2 + r_2^2

and rearranging gives the identity for the sum of squares.

Substitute the known values into the identity and solve for

c

10 = 4^2 - 2\left(\frac{c}{3}\right) \implies 10 = 16 - \frac{2c}{3} \implies \frac{2c}{3} = 6 \implies c = 9

Substituting

r_1^2 + r_2^2 = 10

and

r_1 + r_2 = 4

isolates the single variable

c

Anahtar Kavram

Vieta's Formulas and Algebraic Identities for Quadratic Equations

Alternatif Yöntem

Alternatively, factor

3(x^2 - 4x + 3) = 0

directly once

r_1

and

r_2

are identified. Since

r_1 + r_2 = 4

and

r_1^2 + r_2^2 = 10

, solving the system of equations for

r_1

and

r_2

yields roots of 1 and 3. The product of these roots is

(1)(3) = 3

, so

\frac{c}{3} = 3

, which gives

c = 9

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