132k views
3 votes
Which systems of equations have no real number solutions? Check all that apply.

y = x2 + 4x + 7 and y = 2
y = x2 – 2 and y = x + 5
y = –x2 – 3 and y = 9 + 2x
y = –3x – 6 and y = 2x2 – 7x
y = x2 and y = 10 – 8x

2 Answers

5 votes

Answer:

A, C, D

Explanation:

I took the test. The other persons was very long so i figured id help make it shorter and straight to the point. Hope this helps :)

User Capser
by
7.0k points
3 votes

QUESTION 1

The first system is


y = {x}^(2) + 4x + 7

and


y = 2

Let us equate the two system to obtain,


{x}^(2) + 4x + 7 = 2

Let us rewrite in general quadratic format to get


{x}^(2) + 4x + 7 - 2 = 0


{x}^(2) + 4x + 5 = 0

This implies that,


a=1,b=4, c=5

We now find the discriminant,


D=b^2-4ac


D=4^2-4(1)(5) = - 4

Since the discriminant is negative, the system has no real number solution.

QUESTION 2

The second system is


y = {x}^(2) - 2

and


y = x + 5

We equate the two system to obtain,


{x}^(2) - 2 = x + 5

This implies that,


{x}^(2) - x - 2 - 5= 0


{x}^(2) - x - 7 = 0


a=1,b=-1,c=-7


D = {b}^(2) - 4ac

We substitute the values to obtain,


D = {( - 1)}^(2) - 4(1)( - 7) = 29

Since the discriminant is positive, the system has real number solutions.

QUESTION 3

The given system is


y = - {x}^(2) - 3


y = 9 + 2x

Equate the system to get,


- {x}^(2) - 3 = 9 + 2x

Rewrite in general quadratic format


- {x}^(2) - 2x - 3 - 9= 0


- {x}^(2) - 2x - 12= 0

Divide through by -1 to get,


{x}^(2) + 2x + 12 = 0


</p><p>a=1,b=2,c=12

Using the discriminant we obtain,


D= {2}^(2) - 4(1)(12) = - 44

Since the discriminant is negative the system has no real number solution.

QUESTION 4

The given system is,


y = - 3x - 6

and


y = 2 {x}^(2) - 7x

Equate the two equations to get,


2 {x}^(2) - 7x = - 3x - 6

Rewrite in general quadratic format, to obtain,


2 {x}^(2) - 7x + 3x + 6 = 0


2 {x}^(2) - 4x + 6 = 0

Divide through by 2 to obtain,


{x}^(2) - 2x + 3= 0

This implies that,


a=1, b=-2,c=3

We calculate the discriminant to obtain,


</p><p>D= {( - 2)}^(2) - 4(1)(3) = - 8

Since the discriminant is negative the system has no real number solution.

QUESTION 5

The given system is


y = {x}^(2)

and


y = 10 -8 x

Equate the two equations to get,


{x}^(2) = 10 - 8x

Rewrite in general quadratic format to obtain,


{x}^(2) + 8x - 10 = 0


a=1,b=8,c=-10

We now calculate the discriminant to get,


D= {8}^(2) - 4(1)( - 10) =104

Since the discriminant is positive the system has real number solutions.

User WBT
by
5.8k points