Consider the equation (x-a)^2 – b = 0, where a and b are both positive real numbers. Write a statement about the number of real solutions for the equation that must be true.
The Correct Answer and Explanation is:
Correct Answer: The equation has two real solutions.
Explanation
Consider the given equation:
(x−a)2−b=0(x – a)^2 – b = 0(x−a)2−b=0
To solve for xxx, we first isolate the squared term:(x−a)2=b(x – a)^2 = b(x−a)2=b
Since both aaa and bbb are positive real numbers, bbb is greater than zero. This means the right-hand side of the equation is a positive number. Taking the square root of both sides:x−a=±bx – a = \pm\sqrt{b}x−a=±b
This gives two distinct solutions:x=a+bandx=a−bx = a + \sqrt{b} \quad \text{and} \quad x = a – \sqrt{b}x=a+bandx=a−b
These are two different real numbers because b\sqrt{b}b is a positive number. Therefore, a+ba + \sqrt{b}a+b and a−ba – \sqrt{b}a−b are not equal.
This equation represents a horizontal shift of a basic quadratic function. The graph of (x−a)2(x – a)^2(x−a)2 is a parabola that opens upward and has its vertex at x=ax = ax=a. When we subtract bbb, the entire graph shifts downward by bbb units. Because bbb is positive, the vertex drops below the x-axis, ensuring that the parabola intersects the x-axis at two distinct points. These points correspond to the two real solutions we found earlier.
This is an important feature of quadratic equations:
- If the expression inside the square equals a positive number, it will have two real solutions.
- If it equals zero, it will have one real solution.
- If it equals a negative number, there are no real solutions.
Since b>0b > 0b>0, the equation always has two real solutions for all positive real values of aaa and bbb.
