Question

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

Oy=x² + 4x + 7 and y = 2
□ y=x²-2 and y=x+5
-Oy=-x²-3 and y = 9 + 2x
y=-3x-6 and y = 2x² - 7x
y=x² and y = 10 - 8x

Answer 1

The first system of equations,
`\( y = x^2 + 4x + 7 \)` and
`\( y = 2 \)`, has no real number solutions.Therefore, option A is correct

The question asks which systems of equations have no real number solutions. To determine this, we can set the quadratic equation equal to the linear equation and look at the discriminant of the resulting quadratic equation. The discriminant,
`D`, is found from the standard quadratic equation
`\( ax^2 + bx + c = 0 \)` and is given by
`\( D = b^2 - 4ac \)`.

If
`\( D < 0 \)`, the quadratic equation has no real solutions.

Let's go through each system and apply this:

1.
`\( y = x^2 + 4x + 7 \)` and
`\( y = 2 \)`

Substitute
`\( y \)` from the second equation into the first:

`\( 2 = x^2 + 4x + 7 \)`

The quadratic equation becomes
`\( x^2 + 4x + 5 = 0 \)`

Here,
`\( a = 1 \)`,
`\( b = 4 \)`, and
`\( c = 5 \)`.

Calculate the discriminant
`\( D = b^2 - 4ac \).`

2.
`\( y = x^2 - 2 \)` and
`\( y = x + 5 \)`

Substitute
`\( y \)` from the second equation into the first:

`\( x + 5 = x^2 - 2 \)`

The quadratic equation becomes
`\( x^2 - x - 7 = 0 \)`

Here,
`\( a = 1 \)`,
`\( b = -1 \)`, and
`\( c = -7 \).`

Calculate the discriminant
`\( D = b^2 - 4ac \).`

3.
`\( y = -x^2 - 3 \)` and
`\( y = 9 + 2x \)`

Substitute \( y \) from the second equation into the first:

`\( 9 + 2x = -x^2 - 3 \)`

The quadratic equation becomes
`\( x^2 + 2x - 12 = 0 \)`

Here,
`\( a = 1 \)`,
`\( b = 2 \)`, and
`\( c = -12 \)`.

Calculate the discriminant
`\( D = b^2 - 4ac \).`

4.
`\( y = -3x - 6 \)` and
`\( y = 2x^2 - 7x \)`

Substitute
`\( y \)` from the second equation into the first:

`\( -3x - 6 = 2x^2 - 7x \)`

The quadratic equation becomes \( 2x^2 - 4x - 6 = 0 \)

Here,
`\( a = 2 \)`,
`\( b = -4 \)`, and
`\( c = -6 \).`

Calculate the discriminant
`\( D = b^2 - 4ac \).`

5.
`\( y = x^2 \)` and
`\( y = 10 - 8x \)`

Substitute \( y \) from the second equation into the first:

`\( 10 - 8x = x^2 \)`

The quadratic equation becomes
`\( x^2 + 8x - 10 = 0 \)`

Here,
`\( a = 1 \)`,
`\( b = 8 \)`, and
`\( c = -10 \).`

Calculate the discriminant
`\( D = b^2 - 4ac \).`

Now we'll calculate the discriminant for each and determine if any are less than zero.

After calculating the discriminant for each system of equations:

1. For the system
`\( y = x^2 + 4x + 7 \)` and
`\( y = 2 \)`, the discriminant is
`\( -4 \)`. Since the discriminant is less than zero, this system has no real number solutions.

2. For the system
`\( y = x^2 - 2 \)` and
`\( y = x + 5 \)`, the discriminant is
`\( 29 \)`. Since the discriminant is greater than zero, this system has real number solutions.

3. For the system
`\( y = -x^2 - 3 \)` and
`\( y = 9 + 2x \)`, the discriminant is
`\( 52 \)`. Since the discriminant is greater than zero, this system has real number solutions.

4. For the system
`\( y = -3x - 6 \)` and
`\( y = 2x^2 - 7x \)`, the discriminant is
`\( 64 \)`. Since the discriminant is greater than zero, this system has real number solutions.

5. For the system
`\( y = x^2 \)` and
`\( y = 10 - 8x \)`, the discriminant is
`\( 104 \)` . Since the discriminant is greater than zero, this system has real number solutions.

Therefore, only the first system of equations,
`\( y = x^2 + 4x + 7 \)` and
`\( y = 2 \)`, has no real number solutions.

Answer 2

Answer + Step-by-step explanation:

Recall That the number of solution

of a quadratic equation ax² + bx + c = 0

depends on the discriminant b² - 4ac :

if b² - 4ac > 0 , the equation has two distinct solutions.

if b² - 4ac = 0 , the equation has only one solution.

if b² - 4ac < 0 , the equation has no solutions.

=======================================

System 1 :

y = x² + 4x + 7 and y = 2

⇔ x² + 4x + 7 = 2

⇔ x² + 4x + 5 = 0

→ b² - 4ac = 4² - 4×1×5 = 16 - 20 = -4 < 0

Then the quadratic equation has no solutions

Therefore the system has no solutions.

System 2 :

y = x² - 2 and y = x + 5

⇔ x² - 2 = x + 5

⇔ x² - x - 7 = 0

→ b² - 4ac = (-1)² + 4×7 = 29 > 0

Then the quadratic equation has two solutions

Therefore the system has two solutions.

System 3 :

y = -x² - 3 and y = 9 + 2x

⇔ -x² - 3 = 9 + 2x

⇔ -x² - 2x - 12 = 0

→ b² - 4ac = (-2)² - 4×(-1)×(-12) = 4 - 48 = -44 < 0

Then the quadratic equation has no solutions

Therefore the system has two solutions.

System 4 :

y = -3x - 6 and y = 2x² - 7x

⇔ -3x - 6 = 2x² - 7x

⇔ 2x² - 4x + 6 = 0

→ b² - 4ac = (-4)² - 4×(2)×(6) = 16 - 48 = -32 < 0

Then the quadratic equation has no solutions

Therefore the system has two solutions.

System 5 :

y = x² and y = 10 - 8x

⇔ x² = 10 - 8x

⇔ x² + 8x - 10 = 0

→ b² - 4ac = 8² - 4×1×(-10) = 64 + 40 = 104 > 0

Then the quadratic equation has two solutions

Therefore the system has two solutions.