In the xy-plane, the system of equations below has exactly one real solution. y = -x^2 + 6… — SAT

In the $xy$ -plane, the system of equations below has exactly one real solution.

y = -x^2 + 6x + 2

y + 2x = k

If $k$ is a constant, what is the value of $k$ ?

Answer: 18

To find the value of the constant

k

for which the system of equations has exactly one real solution, we can solve the system by substitution. Substituting

y = k - 2x

from the second equation into the first equation gives

k - 2x = -x^2 + 6x + 2

. Rearranging the terms to write this quadratic equation in standard form,

ax^2 + bx + c = 0

, yields

x^2 - 8x + (k - 2) = 0

. A quadratic equation has exactly one real solution when its discriminant,

b^2 - 4ac

, is equal to zero. Substituting

a = 1

b = -8

, and

c = k - 2

into the discriminant formula gives

(-8)^2 - 4(1)(k - 2) = 0

. Simplifying this expression results in

64 - 4k + 8 = 0

, which simplifies further to

72 - 4k = 0

. Solving for

k

yields

k = 18

Rewrite the linear equation to express

y

in terms of

x

and

k

y = k - 2x

This allows substituting the expression for

y

into the quadratic equation to eliminate

y

Substitute the expression for

y

into the quadratic equation.

k - 2x = -x^2 + 6x + 2

To construct a single quadratic equation in terms of

x

whose solutions correspond to the system's intersection points.

Rearrange the equation into standard quadratic form

ax^2 + bx + c = 0

x^2 - 8x + (k - 2) = 0

To identify the coefficients

a = 1

b = -8

, and

c = k - 2

Set the discriminant of the quadratic equation to zero.

(-8)^2 - 4(1)(k - 2) = 0

A quadratic equation has exactly one real solution if and only if its discriminant is zero.

Solve the linear equation for

k

k = 18

Simplify the equation:

64 - 4k + 8 = 0 \implies 72 - 4k = 0 \implies 4k = 72 \implies k = 18

Nonlinear Systems of Equations

Question