Question

Jordan is solving this system of equations: y = 2x2 + 3 and y – x = 6. Which statements are true about Jordan’s system? Check all that apply. The quadratic equation is in standard form. Using substitution, the system of equations can be rewritten as 2x2 – x – 3 = 0. There are two real number solutions. There are no real number solutions. A solution of the system of equations is (–1, 1.5).

Answer 1

The system of equations provided by Jordan can be solved using the quadratic formula. There are two real number solutions for x, which can then be used to find the corresponding values of y. The given solution of (-1, 1.5) does not satisfy the second equation of the system.

Step-by-step explanation:

The system of equations provided by Jordan can be rewritten as:

y = 2x2 + 3

y - x = 6

To solve the system, we can use the substitution method. We substitute the value of y in the second equation with 2x2 + 3:

2x2 + 3 - x = 6

2x2 - x - 3 = 0

This quadratic equation is not in standard form. The correct form is:

2x2 - x - 3 = 0

Using the quadratic formula, we can find the solution for x:

x = (-b ± √(b² - 4ac)) / (2a)

For 2x2 - x - 3 = 0, a = 2, b = -1, and c = -3.

Substituting these values into the quadratic formula:

x = (-(-1) ± √((-1)² - 4(2)(-3))) / (2(2))

x = (1 ± √(1 + 24)) / 4

x = (1 ± √25) / 4

There are two real number solutions for x: x = (1 + √25) / 4 and x = (1 - √25) / 4.

Now we can find the corresponding values of y using the second equation:

y - x = 6

y - (1 + √25) / 4 = 6

y = (1 + √25) / 4 + 6

y = (1 + √25 + 24) / 4

y = (25 + 1 + √25) / 4

y = (26 + √25) / 4

Similarly, we can find the value of y for the other solution of x. However, the given solution of (-1, 1.5) does not satisfy the second equation of the system, so it is not a solution.

Answer 2

• The quadratic equation is in standard form.

• Using substitution, the system of equations can be rewritten as 2x2 – x – 3 = 0.

• There are two real number solutions.