91Ó°ÊÓ

To solve the equation, we use the concept of the square root. The square root of a number is another number that, when multiplied by itself, gives the original number. For example, the square root of 9 is 3 because: \ 3 \times 3 = 9.

In this exercise, we need to take the square root of both sides of the equation: \ x^2 = 4b. \ Taking the square root on both sides isolates the variable x: \ \ \ \( \sqrt{x^2} = \sqrt{4b} \) \ \ The square root and the square cancel each other on the left side, hence: \ \ \( x = \pm \sqrt{4b} \) \ \ The \( \pm \) symbol means that there are two solutions: one positive and one negative.

However, since we are asked to assume that a and b are positive real numbers, it simplifies our discussion on positive solutions.

Simplifying equations is a critical part of solving quadratic equations. It involves reducing an equation to its simplest form, making it easier to find the solution.

After taking the square root in the previous step, we obtained: \ \( x = \pm \sqrt{4b} \) \ Let's simplify \( \sqrt{4b} \): \ \ We know \( 4 = 2^2 \), so:

\ \( \sqrt{4b} = \sqrt{2^2b} \) \ \ We can separate the constant and the variable under the radical:

\ \( \sqrt{2^2b} = \sqrt{2^2} \cdot \sqrt{b} \) \ \ Which simplifies further to: \ \ \( 2 \sqrt{b} \) \ Finally, we substitute this back into the equation:

\ \( x = \pm 2\sqrt{b} \)

The exercise mentions that we assume a and b represent positive real numbers. Understanding positive real numbers is essential in this context. Positive real numbers are all the numbers greater than zero.

Examples include: \ - Natural numbers like 1, 2, 3, etc. \ - Fractions and decimals like 1/2, 0.75, \ \ This assumption simplifies our solution. Since \( b \) is a positive real number, \ \( 2\sqrt{b} \) will always produce a real number.

When we encounter square roots of positive real numbers, we always consider the principal square root, which is the non-negative root.

Thus, in this case, simplifying the square root of a positive real number ensures that we get a meaningful, valid solution. \ So, our final solution for \( x \) considers both \( x = 2 \sqrt{b} \) and \( x = -2 \sqrt{b} \).