\begin{aligned} y &= x^2 - 8x + c \\ y &= 2x - 5 \end{aligned} In the system of equations … — SAT

\begin{aligned} y &= x^2 - 8x + c \\ y &= 2x - 5 \end{aligned}

In the system of equations above, $c$ is a constant. If the system has two real solutions, $(x_1, y_1)$ and $(x_2, y_2)$ , such that the product of the $y$ -coordinates of the solutions, $y_1 y_2$ , is equal to $5$ , what is the value of $c$ ?

Cevap: 15

Cevap

The value of the constant

c

15

Substituting the expression for

y

from the linear equation into the quadratic equation yields the single variable quadratic equation

x^2 - 10x + (c+5) = 0

. Using Vieta's formulas, the sum of the roots is

x_1 + x_2 = 10

and the product of the roots is

x_1 x_2 = c+5

. Substituting these relationships into the expanded product of the

y

-coordinates,

y_1 y_2 = (2x_1 - 5)(2x_2 - 5) = 4x_1 x_2 - 10(x_1 + x_2) + 25

, allows us to set up the equation

5 = 4(c+5) - 100 + 25

. Solving for

c

yields

15

. Checking the discriminant of the quadratic equation at

c=15

gives

100 - 80 = 20

, which is positive, confirming the existence of two distinct real solutions.

Adım Adım Çözüm

Substitute the expression for

y

from the second equation into the first equation.

2x - 5 = x^2 - 8x + c \implies x^2 - 10x + (c+5) = 0

This substitution reduces the system to a single quadratic equation whose roots,

x_1

and

x_2

, represent the

x

-coordinates of the intersection points.

Apply Vieta's formulas to the resulting quadratic equation.

x_1 + x_2 = 10

and

x_1 x_2 = c+5

Vieta's formulas state that for a quadratic equation

ax^2 + bx + d = 0

, the sum of the roots is

-\frac{b}{a}

and the product of the roots is \frac{d}{a}.

Express the product of the

y

-coordinates,

y_1 y_2

, in terms of

x_1

and

x_2

using the linear relationship.

y_1 y_2 = (2x_1 - 5)(2x_2 - 5) = 4x_1 x_2 - 10(x_1 + x_2) + 25

Since both intersection points lie on the line

y = 2x - 5

, we can substitute

y_1 = 2x_1 - 5

and

y_2 = 2x_2 - 5

and expand.

Substitute the Vieta's formulas relations into the product equation and solve for

c

5 = 4(c + 5) - 10(10) + 25 \implies 5 = 4c + 20 - 100 + 25 \implies 5 = 4c - 55 \implies 60 = 4c \implies c = 15

By substituting the known values of

(x_1 + x_2)

and

(x_1 x_2)

and setting the product

y_1 y_2

5

, we obtain a linear equation in terms of

c

that we can solve directly.

Anahtar Kavram

Solving systems of linear-quadratic equations using algebraic substitution and Vieta's formulas.

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