If the quadratic equation 2x^2 - kx + 18 = 0 has two positive real roots x_1 and x_2 such … — GRE General Test

If the quadratic equation $2x^2 - kx + 18 = 0$ has two positive real roots $x_1$ and $x_2$ such that $x_1 = 4x_2$ , what is the value of the constant $k$ ?

$15$ Cevap
B
$\frac{15}{2}$
C
$30$
D
$9$
E
$24$

Cevap

15

By Vieta's formulas for

ax^2 + bx + c = 0

, the product of the roots is

x_1 x_2 = \frac{c}{a} = \frac{18}{2} = 9

. Given

x_1 = 4x_2

, substituting gives

4x_2^2 = 9

, which yields

x_2 = \frac{3}{2}

since roots are positive. Then

x_1 = 6

. The sum of the roots is

x_1 + x_2 = \frac{15}{2}

. By Vieta's formulas,

x_1 + x_2 = -\frac{b}{a} = \frac{k}{2}

. Setting

\frac{k}{2} = \frac{15}{2}

gives

k = 15

Adım Adım Çözüm

Apply Vieta's formula for the product of the roots

x_1 \cdot x_2 = \frac{18}{2} = 9

For a quadratic equation

ax^2 + bx + c = 0

, the product of the roots is

\frac{c}{a}

Substitute the given relationship

x_1 = 4x_2

into the product equation

(4x_2)(x_2) = 9 \implies 4x_2^2 = 9 \implies x_2^2 = \frac{9}{4} \implies x_2 = \frac{3}{2}

Since the roots are positive, we take the positive square root of

\frac{9}{4}

Calculate the larger root

x_1

x_1 = 4 \left(\frac{3}{2}\right) = 6

Using the relationship

x_1 = 4x_2

Apply Vieta's formula for the sum of the roots to solve for

k

x_1 + x_2 = 6 + \frac{3}{2} = \frac{15}{2} = \frac{k}{2} \implies k = 15

For

ax^2 + bx + c = 0

, the sum of the roots is

-\frac{b}{a} = \frac{k}{2}

Anahtar Kavram

Relating roots of a quadratic equation to its coefficients using Vieta's formulas and factoring relationships.

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