In the quadratic equation x^2 - bx + 64 = 0, b is a constant. The two real solutions to th… — SAT

In the quadratic equation $x^2 - bx + 64 = 0$ , $b$ is a constant. The two real solutions to the equation are $r$ and $s$ , where $r = s^2$ . What is the value of $b$ ?

Cevap: 20

Cevap

By applying Vieta's formulas to the quadratic equation

x^2 - bx + 64 = 0

, we find that the product of the solutions is

r \cdot s = 64

. Substituting the given relation

r = s^2

gives

s^3 = 64

, which yields the real solution

s = 4

. Substituting this back into the relation gives the other solution

r = 16

. Finally, the sum of the solutions is

r + s = b

, so

b = 16 + 4 = 20

Adım Adım Çözüm

Apply Vieta's formula for the product of the roots

r \cdot s = 64

For a quadratic equation in the form

x^2 - bx + c = 0

, the product of the roots is equal to the constant term

c

Substitute the given root relationship into the product equation

s^3 = 64

, which solves to

s = 4

We are given that one root is the square of the other (

r = s^2

), so substituting

s^2

for

r

allows us to solve for

s

Calculate the value of the second root

r

r = 16

Using the relation

r = s^2

with

s = 4

, we find

r = 4^2 = 16

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

b

b = 20

For the equation

x^2 - bx + 64 = 0

, the sum of the roots is

r + s = b

. Substituting

r = 16

and

s = 4

gives

16 + 4 = 20

Anahtar Kavram

Vieta's formulas relating the coefficients of a quadratic equation to its roots

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