In the quadratic equation 2x^2 - bx + 18 = 0, b is a positive constant. If one of the solu… — SAT

In the quadratic equation $2x^2 - bx + 18 = 0$ , $b$ is a positive constant. If one of the solutions to the equation is $4$ times the other solution, what is the value of $b$ ?

Answer: 15

Answer

By Vieta's formulas, the product of the roots of the quadratic equation

2x^2 - bx + 18 = 0

\frac{18}{2} = 9

. Letting the roots be

r_1

and

r_2

with

r_1 = 4r_2

, we have

4r_2^2 = 9

, which yields

r_2 = \frac{3}{2}

(since

b > 0

implies the roots must be positive). Thus,

r_1 = 6

. The sum of the roots is

6 + \frac{3}{2} = \frac{15}{2}

, and by Vieta's formulas, this sum equals

\frac{b}{2}

. Solving for

b

gives

15

Step-by-Step Solution

Set up the relationships for the product and sum of the roots using Vieta's formulas.

r_1 \cdot r_2 = 9

and

r_1 + r_2 = \frac{b}{2}

Vieta's formulas relate the coefficients of a quadratic equation to the sum and product of its roots.

Substitute the given condition that one root is

4

times the other (

r_1 = 4r_2

) into the product equation.

4r_2^2 = 9

This reduces the product equation to a single variable equation in terms of

r_2

Solve for

r_2

and determine its sign based on the constraint that

b

is positive.

r_2 = \frac{3}{2}

Since

b > 0

, the sum of the roots

5r_2 = \frac{b}{2}

must be positive, which requires

r_2 > 0

Calculate the second root

r_1

and then use the sum of the roots to find

b

b = 15

The sum of the roots is

6 + \frac{3}{2} = \frac{15}{2}

, and since

r_1 + r_2 = \frac{b}{2}

, we have

\frac{b}{2} = \frac{15}{2}

Key Concept

Relationship between roots and coefficients of a quadratic equation

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