Solving the Expression for x

The given problem asks us to evaluate the expression \(bx^2 - ax + b\) where \(x\) is defined by a fraction involving square roots. The expression for \(x\) is:

\[ x = \frac{\sqrt{a+2 b}+\sqrt{a−2 b}}{\sqrt{a+2 b}−\sqrt{a−2 b}} \]

This form of expression \(\frac{\sqrt{A}+\sqrt{B}}{\sqrt{A}−\sqrt{B}}\) is suitable for applying the componendo and dividendo rule. The rule states that if \(\frac{p}{q} = \frac{r}{s}\), then \(\frac{p+q}{p-q} = \frac{r+s}{r-s}\).

Applying Componendo and Dividendo to x

We can write \(x\) as \(\frac{x}{1}\). Let \(p=x\), \(q=1\), \(r=\sqrt{a+2b}+\sqrt{a-2b}\), and \(s=\sqrt{a+2b}-\sqrt{a-2b}\). Applying the componendo and dividendo rule:

\[ \frac{x+1}{x-1} = \frac{(\sqrt{a+2 b}+\sqrt{a−2 b}) + (\sqrt{a+2 b}−\sqrt{a−2 b})}{(\sqrt{a+2 b}+\sqrt{a−2 b}) − (\sqrt{a+2 b}−\sqrt{a−2 b})} \]

Let's simplify the numerator and the denominator of the right side:

• Numerator: \((\sqrt{a+2 b}+\sqrt{a−2 b}) + (\sqrt{a+2 b}−\sqrt{a−2 b}) = \sqrt{a+2 b} + \sqrt{a−2 b} + \sqrt{a+2 b} − \sqrt{a−2 b} = 2\sqrt{a+2 b}\)
• Denominator: \((\sqrt{a+2 b}+\sqrt{a−2 b}) − (\sqrt{a+2 b}−\sqrt{a−2 b}) = \sqrt{a+2 b} + \sqrt{a−2 b} − \sqrt{a+2 b} + \sqrt{a−2 b} = 2\sqrt{a−2 b}\)

So, the equation becomes:

\[ \frac{x+1}{x-1} = \frac{2\sqrt{a+2 b}}{2\sqrt{a−2 b}} = \sqrt{\frac{a+2 b}{a−2 b}} \]

Solving for x and the Expression

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

\[ \left(\frac{x+1}{x-1}\right)^2 = \left(\sqrt{\frac{a+2 b}{a−2 b}}\right)^2 \]

\[ \frac{(x+1)^2}{(x-1)^2} = \frac{a+2 b}{a−2 b} \]

\[ \frac{x^2+2x+1}{x^2-2x+1} = \frac{a+2 b}{a−2 b} \]

Now, we can apply componendo and dividendo again to this equation:

\[ \frac{(x^2+2x+1) + (x^2-2x+1)}{(x^2+2x+1) − (x^2-2x+1)} = \frac{(a+2 b) + (a−2 b)}{(a+2 b) − (a−2 b)} \]

Let's simplify both sides:

• Left Numerator: \((x^2+2x+1) + (x^2-2x+1) = x^2+2x+1+x^2-2x+1 = 2x^2+2\)
• Left Denominator: \((x^2+2x+1) − (x^2-2x+1) = x^2+2x+1-x^2+2x-1 = 4x\)
• Right Numerator: \((a+2 b) + (a−2 b) = a+2b+a-2b = 2a\)
• Right Denominator: \((a+2 b) − (a−2 b) = a+2b-a+2b = 4b\)

Substituting these back into the equation:

\[ \frac{2x^2+2}{4x} = \frac{2a}{4b} \]

\[ \frac{2(x^2+1)}{4x} = \frac{2a}{4b} \]

\[ \frac{x^2+1}{2x} = \frac{a}{2b} \]

Now, we can cross-multiply to find a relationship between \(x\), \(a\), and \(b\):

\[ (x^2+1)(2b) = a(2x) \]

\[ 2b x^2 + 2b = 2ax \]

Divide the entire equation by 2 (since b ≠ 0 and x is involved, it's generally safe):

\[ b x^2 + b = ax \]

Rearrange the terms to get the expression we need to evaluate:

\[ b x^2 − ax + b = 0 \]

Thus, the expression \(bx^2 - ax + b\) is equal to 0.

Result and Conclusion

Based on our simplification of the expression for \(x\) and the subsequent algebraic manipulations, we found that \(bx^2 - ax + b\) is equal to 0.

The final answer is 0.

Revision Table: Key Steps and Formulas

Additional Information on Componendo and Dividendo

The componendo and dividendo rule is a very useful property of proportions. If we have a proportion \(\frac{a}{b} = \frac{c}{d}\), it implies that \(ad = bc\).

• Componendo: If \(\frac{a}{b} = \frac{c}{d}\), then \(\frac{a+b}{b} = \frac{c+d}{d}\). This is derived by adding 1 to both sides: \(\frac{a}{b} + 1 = \frac{c}{d} + 1 \implies \frac{a+b}{b} = \frac{c+d}{d}\).
• Dividendo: If \(\frac{a}{b} = \frac{c}{d}\), then \(\frac{a-b}{b} = \frac{c-d}{d}\). This is derived by subtracting 1 from both sides: \(\frac{a}{b} - 1 = \frac{c}{d} - 1 \implies \frac{a-b}{b} = \frac{c-d}{d}\).
• Componendo and Dividendo: If \(\frac{a}{b} = \frac{c}{d}\), then \(\frac{a+b}{a-b} = \frac{c+d}{c-d}\). This is derived by dividing the componendo result by the dividendo result: \(\frac{(a+b)/b}{(a-b)/b} = \frac{(c+d)/d}{(c-d)/d} \implies \frac{a+b}{a-b} = \frac{c+d}{c-d}\). This rule is particularly effective when dealing with expressions like \(\frac{p+q}{p-q}\) or ratios involving sums and differences.