Solving the Radical Equation using Componendo and Dividendo

We are given the equation \(\rm\frac{\sqrt{x+20}+\sqrt{x-1}}{\sqrt{x+20}-\sqrt{x-1}}=\frac{7}{3}\) and asked to find the value of \(\rm \sqrt{(x + 20)(x-1)}\).

This equation has a structure that is perfectly suited for applying the property of Componendo and Dividendo. The property states that if \(\rm \frac{a}{b} = \frac{c}{d}\), then \(\rm \frac{a+b}{a-b} = \frac{c+d}{c-d}\). However, in our case, the equation is already in the form \(\rm \frac{a+b}{a-b} = \frac{c}{d}\), where \(\rm a = \sqrt{x+20}\) and \(\rm b = \sqrt{x-1}\), and \(\rm \frac{c}{d} = \frac{7}{3}\). We can apply the inverse form of the property or apply it directly as follows:

Given:

\(\rm \frac{\sqrt{x+20}+\sqrt{x-1}}{\sqrt{x+20}-\sqrt{x-1}}=\frac{7}{3}\)

Applying Componendo and Dividendo:

\(\rm \frac{(\sqrt{x+20}+\sqrt{x-1}) + (\sqrt{x+20}-\sqrt{x-1})}{(\sqrt{x+20}+\sqrt{x-1}) - (\sqrt{x+20}-\sqrt{x-1})} = \frac{7+3}{7-3}\)

Simplify the numerator and the denominator on the left side:

• Numerator: \(\rm \sqrt{x+20}+\sqrt{x-1} + \sqrt{x+20}-\sqrt{x-1} = 2\sqrt{x+20}\)
• Denominator: \(\rm \sqrt{x+20}+\sqrt{x-1} - \sqrt{x+20} + \sqrt{x-1} = 2\sqrt{x-1}\)

Simplify the right side:

• \(\rm \frac{7+3}{7-3} = \frac{10}{4} = \frac{5}{2}\)

So the equation becomes:

\(\rm \frac{2\sqrt{x+20}}{2\sqrt{x-1}} = \frac{5}{2}\)

\(\rm \frac{\sqrt{x+20}}{\sqrt{x-1}} = \frac{5}{2}\)

Solving for x in the Simplified Equation

To eliminate the square roots, we square both sides of the equation:

\(\rm \left(\frac{\sqrt{x+20}}{\sqrt{x-1}}\right)^2 = \left(\frac{5}{2}\right)^2\)

\(\rm \frac{x+20}{x-1} = \frac{25}{4}\)

Now, we cross-multiply to solve for \(\rm x\):

\(\rm 4(x+20) = 25(x-1)\)

\(\rm 4x + 80 = 25x - 25\)

Rearrange the terms to group \(\rm x\) terms and constant terms:

\(\rm 80 + 25 = 25x - 4x\)

\(\rm 105 = 21x\)

Divide by 21 to find the value of \(\rm x\):

\(\rm x = \frac{105}{21}\)

\(\rm x = 5\)

Finding the Value of the Expression

We need to find the value of \(\rm \sqrt{(x + 20)(x-1)}\) when \(\rm x = 5\).

Substitute \(\rm x = 5\) into the expression:

\(\rm \sqrt{(5 + 20)(5-1)}\)

\(\rm \sqrt{(25)(4)}\)

\(\rm \sqrt{100}\)

\(\rm 10\)

Thus, the value of \(\rm \sqrt{(x + 20)(x-1)}\) is 10.

Revision Table: Key Concepts

Additional Information: Domain of Radical Equations

When solving radical equations, it's important to consider the domain of the variables. The expressions under a square root must be non-negative.

For \(\rm \sqrt{x+20}\) to be defined, \(\rm x+20 \ge 0 \implies x \ge -20\).

For \(\rm \sqrt{x-1}\) to be defined, \(\rm x-1 \ge 0 \implies x \ge 1\).

For both radicals to be defined, \(\rm x\) must satisfy both conditions, so \(\rm x \ge 1\).

Our solution \(\rm x = 5\) satisfies this condition (\(\rm 5 \ge 1\)), so it is a valid solution.

Also, the denominators in the original equation cannot be zero: \(\rm \sqrt{x+20}-\sqrt{x-1} \ne 0\), which means \(\rm \sqrt{x+20} \ne \sqrt{x-1}\). Squaring both sides gives \(\rm x+20 \ne x-1\), which simplifies to \(\rm 20 \ne -1\). This is always true, so this condition doesn't exclude \(\rm x=5\).