Math Problem Statement

Solve the equation: \(\frac{\sqrt{a + x} + \sqrt{a - x}}{\sqrt{a + x} - \sqrt{a - x}} = \frac{a}{x}\)

Solution

The given equation is:

a+x+axa+xax=ax\frac{\sqrt{a + x} + \sqrt{a - x}}{\sqrt{a + x} - \sqrt{a - x}} = \frac{a}{x}

We can solve this equation by manipulating the left-hand side. Let's start by simplifying the numerator and denominator separately.

Step 1: Multiply by the conjugate

We multiply both the numerator and the denominator by the conjugate of the denominator to eliminate the square roots in the denominator. The conjugate of a+xax\sqrt{a + x} - \sqrt{a - x} is a+x+ax\sqrt{a + x} + \sqrt{a - x}. So:

(a+x+ax)(a+xax)×(a+x+ax)(a+x+ax)\frac{\left( \sqrt{a + x} + \sqrt{a - x} \right)}{\left( \sqrt{a + x} - \sqrt{a - x} \right)} \times \frac{\left( \sqrt{a + x} + \sqrt{a - x} \right)}{\left( \sqrt{a + x} + \sqrt{a - x} \right)}

This simplifies to:

(a+x+ax)2(a+x)2(ax)2\frac{\left( \sqrt{a + x} + \sqrt{a - x} \right)^2}{\left( \sqrt{a + x} \right)^2 - \left( \sqrt{a - x} \right)^2}

Step 2: Simplify the denominator

Using the difference of squares formula, we simplify the denominator:

(a+x)2(ax)2=(a+x)(ax)=a+xa+x=2x\left( \sqrt{a + x} \right)^2 - \left( \sqrt{a - x} \right)^2 = (a + x) - (a - x) = a + x - a + x = 2x

Thus, the denominator becomes 2x2x.

Step 3: Expand the numerator

Now expand the numerator (a+x+ax)2\left( \sqrt{a + x} + \sqrt{a - x} \right)^2:

(a+x+ax)2=(a+x)2+2(a+x)(ax)+(ax)2\left( \sqrt{a + x} + \sqrt{a - x} \right)^2 = (\sqrt{a + x})^2 + 2\sqrt{(a + x)(a - x)} + (\sqrt{a - x})^2

Simplifying:

=(a+x)+2(a+x)(ax)+(ax)= (a + x) + 2\sqrt{(a + x)(a - x)} + (a - x)

The xx's cancel out, leaving:

a+a+2(a+x)(ax)=2a+2a2x2a + a + 2\sqrt{(a + x)(a - x)} = 2a + 2\sqrt{a^2 - x^2}

Step 4: Simplify the full expression

Now, the left-hand side becomes:

2a+2a2x22x\frac{2a + 2\sqrt{a^2 - x^2}}{2x}

We can factor out the 2 in the numerator:

2(a+a2x2)2x=a+a2x2x\frac{2(a + \sqrt{a^2 - x^2})}{2x} = \frac{a + \sqrt{a^2 - x^2}}{x}

Step 5: Compare with the right-hand side

We are given that the right-hand side is ax\frac{a}{x}. So:

a+a2x2x=ax\frac{a + \sqrt{a^2 - x^2}}{x} = \frac{a}{x}

For this to hold true, the term a2x2\sqrt{a^2 - x^2} must be equal to 0, which implies:

a2x2=0a2=x2a=xa^2 - x^2 = 0 \quad \Rightarrow \quad a^2 = x^2 \quad \Rightarrow \quad a = x

Conclusion

The solution to the equation is a=xa = x.


Would you like more details on any step, or do you have any further questions?

Here are 5 related questions to explore:

  1. How do you find the conjugate of a binomial containing square roots?
  2. What is the difference of squares formula, and how is it used?
  3. How do you simplify expressions involving square roots?
  4. Can you solve other equations by multiplying by conjugates?
  5. What happens if you have more terms under the square roots?

Tip: Always check for common patterns like the difference of squares when simplifying complex square root expressions!

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Math Problem Analysis

Mathematical Concepts

Algebra
Radicals
Simplification

Formulas

Difference of squares formula: \(a^2 - b^2 = (a - b)(a + b)\)
Conjugates of radicals

Theorems

Difference of squares
Equality condition for radicals

Suitable Grade Level

Grades 10-12