In the xy-plane, the line y = x - 4 intersects the parabola y = x^2 - 3x - 4 at the points… — SAT

In the $xy$ -plane, the line $y = x - 4$ intersects the parabola $y = x^2 - 3x - 4$ at the points $(x_1, y_1)$ and $(x_2, y_2)$ . If $x_1 < x_2$ , what is the value of $x_2$ ?

A
0
B
2
4Answer
D
-4

Answer

The correct answer is 4. Setting the two equations equal to find their points of intersection gives

x^2 - 3x - 4 = x - 4

. Subtracting

x

and adding

4

to both sides yields the simplified quadratic equation

x^2 - 4x = 0

. Factoring out

x

gives

x(x - 4) = 0

, which has the solutions

x = 0

and

x = 4

. Given that

x_1 < x_2

, we have

x_1 = 0

and

x_2 = 4

. Therefore, the value of

x_2

is 4.

Step-by-Step Solution

Set the two equations equal to each other to find their points of intersection.

x^2 - 3x - 4 = x - 4

At the intersection points, the y-values of the line and the parabola must be equal.

Subtract

x

and add

4

to both sides of the equation to set it to zero.

x^2 - 4x = 0

To solve a quadratic equation, we must rewrite it in standard form:

ax^2 + bx + c = 0

Factor the quadratic expression by factoring out the greatest common factor, which is

x

x(x - 4) = 0

Factoring allows us to use the zero product property to find the individual roots.

Solve for the two possible values of

x

x = 0

x = 4

Setting each factor to zero gives

x = 0

and

x - 4 = 0

, which simplifies to

x = 4

Compare the two solutions to find the value of

x_2

given the condition

x_1 < x_2

x_1 = 0

and

x_2 = 4

Since

0 < 4

, the smaller value is

x_1

and the larger value is

x_2

Key Concept

Solving a system of a linear equation and a quadratic equation by substitution.

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