For the quadratic equation 2x^2 + bx + 16 = 0, where b is a positive constant, the sum of … — ACT

For the quadratic equation $2x^2 + bx + 16 = 0$ , where $b$ is a positive constant, the sum of the squares of the two complex solutions is equal to $-7$ . What is the value of $b$ ?

Cevap: 6

Cevap

The value of the positive constant

b

6

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

2x^2 + bx + 16 = 0

-\frac{b}{2}

and the product is

8

. Using the identity

x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2

, we substitute

-7

for the sum of the squares, yielding

-7 = \left(-\frac{b}{2}\right)^2 - 2(8)

. This simplifies to

-7 = \frac{b^2}{4} - 16

, which leads to

\frac{b^2}{4} = 9

and

b^2 = 36

. Since

b

is a positive constant,

b = 6

Adım Adım Çözüm

Find the sum and product of the roots in terms of

b

using Vieta's formulas.

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

and

x_1 x_2 = 8

Vieta's formulas state that for a 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 the squares of the roots to their sum and product.

x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2

This algebraic identity allows us to express the sum of the squares of the roots using the sum and product of the roots.

Substitute the known values into the identity and solve for

b

-7 = \left(-\frac{b}{2}\right)^2 - 2(8) \implies -7 = \frac{b^2}{4} - 16 \implies \frac{b^2}{4} = 9 \implies b^2 = 36 \implies b = 6

Substituting the given sum of squares (

-7

), sum (

-\frac{b}{2}

), and product (

8

) produces a single-variable equation that can be solved for the positive constant

b

Anahtar Kavram

Using Vieta's formulas and algebraic identities to relate the roots of a quadratic equation to its coefficients.

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