Question

1 determine the number of real solutions of the system below. \

$$\begin{cases} y = 2x^2 + 1 \\\\ y = -x \\end{cases}$$

\text{a) } 2 \\\\ \text{b) } 1 \\\\ \text{c) } 0 \\\\ \text{d) } \text{more than 2}

Explanation:

Step1: Substitute \( y = -x \) into \( y = 2x^2 + 1 \)

Substitute to get the equation: \( -x = 2x^2 + 1 \)

Step2: Rearrange the equation to standard quadratic form

Rearrange the equation: \( 2x^2 + x + 1 = 0 \)

Step3: Calculate the discriminant of the quadratic equation

For a quadratic equation \( ax^2 + bx + c = 0 \), the discriminant \( D = b^2 - 4ac \). Here, \( a = 2 \), \( b = 1 \), \( c = 1 \). So \( D = 1^2 - 4 \times 2 \times 1 = 1 - 8 = -7 \)

Step4: Analyze the discriminant

Since the discriminant \( D = -7 < 0 \), the quadratic equation \( 2x^2 + x + 1 = 0 \) has no real roots. Therefore, the system of equations has no real solutions.

Answer:

C. 0