A system of equations consists of the line 2x - y = 5 and the parabola y = x^2 - 4x + c, w… — ACT

A system of equations consists of the line $2x - y = 5$ and the parabola $y = x^2 - 4x + c$ , where $c$ is a constant. If the line and the parabola intersect at two distinct points $(x_1, y_1)$ and $(x_2, y_2)$ such that the positive difference between their $x$ -coordinates is 2, what is the value of $c$ ?

A
0
3Answer
C
5
D
8
E
11

Answer

To find the constant

c

, we equate the line and the parabola equations:

x^2 - 4x + c = 2x - 5

. Bringing all terms to one side gives the quadratic equation

x^2 - 6x + (c + 5) = 0

. Applying the quadratic formula, the

x

-coordinates of the intersection points are

x = 3 \pm \sqrt{4 - c}

. The positive difference between these coordinates is

(3 + \sqrt{4 - c}) - (3 - \sqrt{4 - c}) = 2\sqrt{4 - c}

. Setting this difference equal to the given value of 2 gives

2\sqrt{4 - c} = 2

, which simplifies to

\sqrt{4 - c} = 1

. Squaring both sides yields

4 - c = 1

, which gives

c = 3

. This corresponds to the correct option.

Step-by-Step Solution

Equate the equations for the line and the parabola to set up the equation for their intersection points.

x^2 - 4x + c = 2x - 5

The intersection points of the system of equations occur where the

y

-values are equal.

Rearrange the equation into standard quadratic form:

ax^2 + bx + c = 0

x^2 - 6x + (c + 5) = 0

Grouping the terms allows us to identify the coefficients:

a = 1

b = -6

, and the constant term is

c + 5

Apply the quadratic formula to find the

x

-coordinates of the intersection points in terms of

c

x = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(c + 5)}}{2} = 3 \pm \sqrt{4 - c}

The quadratic formula yields the roots

x_1 = 3 + \sqrt{4 - c}

and

x_2 = 3 - \sqrt{4 - c}

Set up an equation representing the positive difference between the

x

-coordinates and solve for

c

(3 + \sqrt{4 - c}) - (3 - \sqrt{4 - c}) = 2 \implies 2\sqrt{4 - c} = 2 \implies \sqrt{4 - c} = 1 \implies 4 - c = 1 \implies c = 3

The difference between the two coordinates is given as 2, which allows us to isolate and solve for

c

Key Concept

Solving systems of linear and non-linear equations by finding the intersection of a line and a parabola and using root properties to determine unknown constants

Alternative Method

Instead of using the quadratic formula, you can apply Vieta's formulas. Let the roots of

x^2 - 6x + (c + 5) = 0

x_1

and

x_2

. Vieta's formulas state that

x_1 + x_2 = 6

and

x_1 x_2 = c + 5

. We are given that

|x_1 - x_2| = 2

. Squaring this equation gives

(x_1 - x_2)^2 = 4

. Since

(x_1 - x_2)^2 = (x_1 + x_2)^2 - 4x_1 x_2

, we substitute the known values:

6^2 - 4(c + 5) = 4 \implies 36 - 4c - 20 = 4 \implies 16 - 4c = 4 \implies 4c = 12 \implies c = 3

Estimated Time:2m 0s

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Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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