Question

Which systems of equations have no real number solutions? Check all that apply.

Answer 1

Answer:

A, C, D

Step-by-step 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 :)

Answer 2

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,

${x}^{2} + 2x + 12 = 0$

$a=1,b=2,c=12$

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,

$D= {( - 2)}^{2} - 4(1)(3) = - 8$

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.