If the quadratic equation x^2 - bx + 16 = 0, where b is a positive constant, has two real … — SAT

If the quadratic equation $x^2 - bx + 16 = 0$ , where $b$ is a positive constant, has two real solutions such that one solution is $4$ times the other, what is the value of $b$ ?

Cevap: 10

Cevap

The value of

b

10

By writing the roots as

r

and

4r

, the quadratic equation can be represented as

(x-r)(x-4r) = x^2 - 5rx + 4r^2 = 0

. Comparing this to the given equation

x^2 - bx + 16 = 0

, we establish that

4r^2 = 16

and

b = 5r

. Solving for

r

gives

r^2 = 4

, which means

r = \pm 2

. Since

b

is a positive constant, we select

r = 2

, yielding

b = 5(2) = 10

. Alternatively, using Vieta's formulas, the product of the roots is

r \cdot 4r = 16 \implies 4r^2 = 16 \implies r = \pm 2

, and the sum of the roots is

r + 4r = b \implies 5r = b

. Since

b > 0

, we find

b = 10

Adım Adım Çözüm

Define the roots in terms of a single variable and express the quadratic equation in factored form.

Let the roots be

r

and

4r

. The factored form is

(x - r)(x - 4r) = 0

This allows us to relate the given relationship between the roots to the coefficients of the quadratic equation.

Expand the factored quadratic expression.

x^2 - 5rx + 4r^2 = 0

This transforms the equation into the standard form

x^2 + Bx + C = 0

so we can match coefficients.

Compare the expanded equation with the given equation

x^2 - bx + 16 = 0

4r^2 = 16

and

b = 5r

Matching corresponding coefficients allows us to solve for the unknown variables.

Solve the constant term equation for

r

r = 2

r = -2

Finding the value of

r

is necessary to calculate the value of

b

Calculate the value of

b

using the constraint that

b

is positive.

b = 5(2) = 10

Using the negative root

r = -2

would yield a negative value for

b

(

b = -10

), which violates the constraint that

b

must be a positive constant.

Anahtar Kavram

Relating the roots of a quadratic equation to its coefficients

Tahmini Süre:1m 30s

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