Question

Which equation has exactly ONE solution? 11 of 15 Next ► 11:51 A 8+3x 2 = 13 + x B 3+12x 2 = 6x + 15 C 10+15z 5 = 3x + 10 D 4-20 4 =x-5

Answer 1

To determine which equation has exactly one solution, let's analyze each option step by step:

A) \( 8 + 3x^2 = 13 + x \)

Rearranging gives:
\[ 3x^2 - x - 5 = 0 \]
This is a quadratic equation. To determine the number of solutions, we can use the discriminant \( D = b^2 - 4ac \):
\[ a = 3, b = -1, c = -5 \]
\[ D = (-1)^2 - 4 \cdot 3 \cdot (-5) = 1 + 60 = 61 \]
Since \( D > 0 \), this equation has two solutions.

B) \( 3 + 12x^2 = 6x + 15 \)

Rearranging gives:
\[ 12x^2 - 6x - 12 = 0 \]
Dividing through by 6:
\[ 2x^2 - x - 2 = 0 \]
Let's find the discriminant:
\[ a = 2, b = -1, c = -2 \]
\[ D = (-1)^2 - 4 \cdot 2 \cdot (-2) = 1 + 16 = 17 \]
Since \( D > 0 \), this equation also has two solutions.

C) \( 10 + 15z^5 = 3x + 10 \)

Rearranging gives:
\[ 15z^5 - 3x = 0 \]
To have a unique solution, let's set \( z = 0 \):
\[ 15(0)^5 - 3x = 0 \]
This means:
\[ -3x = 0 \implies x = 0 \]
So it depends on if \( z \) could equal more values. Since \( z^5 \) could take many values, this does not lead to one solution.

D) \( 4 - 20 = x - 5 \)

Simplifying gives:
\[ -16 = x - 5 \]
So,
\[ x = -11 \]
This is indeed a unique solution.

After analyzing all equations, the equation that has exactly one solution is:

**D) \( 4 - 20 = x - 5 \)**