how many real-number solutions does the equation have?
y=x^2 + 2x + 3
a one solution
b two solutions
c no solutions
d infinitely many solutions
mmmmhhh ?????
I apologize for the mistake. The answer is actually c, no solutions. This can be seen by noting that the quadratic y = x^2 + 2x + 3 opens upward (since its leading coefficient is positive), and therefore its graph never intersects the x-axis, which means there are no real-number solutions to the equation y = 0.
Answer is true, but not for the reasons you state.
there are many quadratic functions I could form that open upwards
and have a leading coefficient of 1 , that will have 1 or 2 solutions
e.g. y = x^2 - x - 12 <--- 2 reals
y = x^2 + 8x + 16 <----- 1 root
You are correct. I apologize for my mistake once again. The correct reason why the equation has no real-number solutions is that the discriminant of the quadratic expression is negative:
b^2 - 4ac = 2^2 - 4(1)(3) = -8.
Since the discriminant is negative, the quadratic has no real-number solutions, which means that the equation y = x^2 + 2x + 3 has no real-number solutions.
To determine the number of real-number solutions for the equation y = x^2 + 2x + 3, we can use the discriminant (Denoted as "D") of the quadratic equation.
The discriminant is calculated as follows: D = b^2 - 4ac, where a, b, and c are the coefficients of the equation in the form ax^2 + bx + c.
In our equation, a = 1, b = 2, and c = 3. Substituting these values into the discriminant formula, we have D = (2^2) - 4(1)(3) = 4 - 12 = -8.
Now, based on the value of the discriminant, we can determine the number of real-number solutions:
1. If the discriminant is greater than zero (D > 0), then the equation has two distinct real solutions.
2. If the discriminant is equal to zero (D = 0), then the equation has one real solution.
3. If the discriminant is less than zero (D < 0), then the equation has no real solutions.
4. If the equation is linear (a = 0) and the constant term (c) is zero, then there are infinitely many solutions.
In our case, the discriminant is negative (D < 0), specifically D = -8. Therefore, the equation y = x^2 + 2x + 3 has no real-number solutions (c).