Question

consider the system of equations below, and find values for m and b so that the system has two solutions
\

$$\begin{cases} y = x^2 \\ y = mx + b \\end{cases}$$

choose the correct answer below
\bigcirc a. \\ m = 1, \\ b = -4
\bigcirc b. \\ m = -2, \\ b = -3
\bigcirc c. \\ m = 0, \\ b = 0
\bigcirc d. \\ m = 1, \\ b = 2

Explanation:

Step1: Substitute \( y = mx + b \) into \( y = x^2 \)

Substitute to get the quadratic equation: \( x^2 = mx + b \), which rearranges to \( x^2 - mx - b = 0 \). For the system to have two solutions, the discriminant of this quadratic (\( \Delta = m^2 + 4b \)) must be greater than 0 (\( \Delta > 0 \)).

Step2: Test Option A (\( m = 1, b = -4 \))

Calculate discriminant: \( \Delta = 1^2 + 4(-4) = 1 - 16 = -15 \). Since \( -15 < 0 \), no real solutions. Eliminate A.

Step3: Test Option B (\( m = -2, b = -3 \))

Calculate discriminant: \( \Delta = (-2)^2 + 4(-3) = 4 - 12 = -8 \). Since \( -8 < 0 \), no real solutions. Eliminate B.

Step4: Test Option C (\( m = 0, b = 0 \))

Equation becomes \( x^2 = 0 \), discriminant \( \Delta = 0^2 + 4(0) = 0 \). Only one solution (a repeated root). Eliminate C.

Step5: Test Option D (\( m = 1, b = 2 \))

Equation becomes \( x^2 - x - 2 = 0 \). Discriminant \( \Delta = 1^2 + 4(2) = 1 + 8 = 9 \). Since \( 9 > 0 \), two real solutions.

Answer:

D. \( m = 1, b = 2 \)