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 = - QAmmunity.org

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 =

asked May 9, 2020 132k views

2 Answers

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.

WBT answered May 14, 2020

by WBT

5.8k points

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