Question

1. error analysis describe and correct the error a student made in solving the system of equations.

y = 2x² + 3
y = 3x + 1

y = 2x² + 3

• (y = 3x + 1)

0 = -x² + 2
x² = 2
x = ±√2 ≈ ±1.41
×

Explanation:

Step1: Identify the error

When subtracting the equations \( y = 2x^2 + 3 \) and \( y=3x + 1 \), the student made a mistake in the coefficient of \( x^2 \) and the linear term. The correct subtraction should be:
\( (2x^2 + 3)-(3x + 1)=0 \)
Expanding this, we get \( 2x^2-3x + 2 = 0 \), not \( 0=-x^2 + 2 \). The student incorrectly subtracted the \( x^2 \) term and missed the linear term (\( - 3x \)).

Step2: Correct the system solution

Set the two equations equal to each other since \( y=y \):
\( 2x^2+3 = 3x + 1 \)
Rearrange to standard quadratic form \( ax^2+bx + c = 0 \):
\( 2x^2-3x+3 - 1=0 \)
\( 2x^2-3x + 2 = 0 \)

Now, use the quadratic formula \( x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \) where \( a = 2 \), \( b=-3 \), and \( c = 2 \).

First, calculate the discriminant \( \Delta=b^2-4ac=(-3)^2-4\times2\times2=9 - 16=-7 \)

Since the discriminant \( \Delta=-7<0 \), the quadratic equation \( 2x^2-3x + 2 = 0 \) has no real solutions. So the system of equations \(

$$\begin{cases}y = 2x^2+3\\y=3x + 1\end{cases}$$

\) has no real - valued solutions for \( x \) (and thus no real - valued solutions for the system).

Answer:

The student made an error in subtracting the two equations (incorrectly handled the \( x^2 \) coefficient and missed the linear term). The correct quadratic equation after setting the two equations equal is \( 2x^2-3x + 2 = 0 \), and since its discriminant \( \Delta=-7<0 \), the system has no real solutions.