Question

Which equation has no real solutions?

Answer 1

3. 4x^2+36 =0 only has a complex solution!

Explanation:

1. Solve for x over the real numbers:

4 x^2 + 4 x - 36 = 0

Divide both sides by 4:

x^2 + x - 9 = 0

Add 9 to both sides:

x^2 + x = 9

Add 1/4 to both sides:

x^2 + x + 1/4 = 37/4

Write the left hand side as a square:

(x + 1/2)^2 = 37/4

Take the square root of both sides:

x + 1/2 = sqrt(37)/2 or x + 1/2 = -sqrt(37)/2

Subtract 1/2 from both sides:

x = sqrt(37)/2 - 1/2 or x + 1/2 = -sqrt(37)/2

Subtract 1/2 from both sides:

Answer: x = sqrt(37)/2 - 1/2 or x = -1/2 - sqrt(37)/2

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2. Solve for x over the real numbers:

4 x^2 - 4 x - 36 = 0

Divide both sides by 4:

x^2 - x - 9 = 0

Add 9 to both sides:

x^2 - x = 9

Add 1/4 to both sides:

x^2 - x + 1/4 = 37/4

Write the left hand side as a square:

(x - 1/2)^2 = 37/4

Take the square root of both sides:

x - 1/2 = sqrt(37)/2 or x - 1/2 = -sqrt(37)/2

Add 1/2 to both sides:

x = 1/2 + sqrt(37)/2 or x - 1/2 = -sqrt(37)/2

Add 1/2 to both sides:

Answer: x = 1/2 + sqrt(37)/2 or x = 1/2 - sqrt(37)/2

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3. Solve for x:

4 (x^2 + 9) = 0

Divide both sides by 4:

x^2 + 9 = 0

Subtract 9 from both sides:

x^2 = -9

Take the square root of both sides:

Answer: x = 3 i or x = -3 i

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4. Solve for x over the real numbers:

4 x^2 - 36 = 0

Add 36 to both sides:

4 x^2 = 36

Divide both sides by 4:

x^2 = 9

Take the square root of both sides:

Answer: x = 3 or x = -3

Answer 2

Among the given quadratic equations, only "4x^2 + 36 = 0" lacks real solutions due to a negative discriminant, while the others possess two real solutions each.

The discriminant, denoted by Δ, is a key factor in determining the nature of solutions to a quadratic equation. For the quadratic equation "ax^2 + bx + c = 0," the discriminant is given by "Δ = b^2 - 4ac."

"4x^2 + 4x - 36 = 0":

Here, a = 4, b = 4, and c = -36. Calculating the discriminant, we get "Δ = 4^2 - 4(4)(-36) = 160." Since Δ > 0, this equation has two real solutions.

"4x^2 - 4x - 36 = 0":

In this case, a = 4, b = -4, and c = -36. The discriminant is "Δ = (-4)^2 - 4(4)(-36) = 256." As Δ > 0, this equation also has two real solutions.

"4x^2 + 36 = 0":

Here, a = 4, b = 0, and c = 36. The discriminant is "Δ = 0^2 - 4(4)(36) = -576." Since Δ < 0, this equation has no real solutions.

"4x^2 - 36 = 0":

In this case, a = 4, b = 0, and c = -36. The discriminant is "Δ = 0^2 - 4(4)(-36) = 576." Like the first two equations, this one has two real solutions.

Therefore, the equation with no real solutions is "4x^2 + 36 = 0."