Try to solve the system by yourself. Analyze your solution to decide where your classmate might have made a mistake.
See solution.
Practice makes perfect
We will start by solving the system ourselves. Let's use the Substitution Method.
y=x^2+2x+4 & (I) y=x+1 & (II)
We will substitute y=x+1 into the first equation and simplify.
We want to solve the first equation for x. To do so, we have to factor it and then use the Zero-Product Property.
x^2+x+3=0 ⇔ 1x^2+1x+3=0
To factor the expression on the left-hand side, we have to find a pair of factors of 3 that has a sum of 1. However, there is only one pair of factors of 3.
Factors of 3: & 1and3
Sum of Factors: & 1+3=4
As we can see the sum of this one pair of factors is not equal to 1. It means that the expression cannot be factorized. Therefore, we have to look for the solutions in a different way. Let's use the Quadratic Formula.
x=- b±sqrt(b^2-4ac)/2a
In our case a=1, b=1, and c=3.
Since the discriminant, b^2-4ac= - 11, is less than 0, the equation x^2+x+3=0 has no solutions. This implies that the whole system has no solutions. Below we have included a graph representing our system.
The classmate might have made an error when trying to factor the first equation. Or they might have made an error when calculating the discriminant in the Quadratic Formula.
