Question

If the ratio of the roots of the equation \(x^2 - 2ax + b = 0\) is equal to that of the roots of the equation \(x^2 - 2cx + d = 0\), then:

• A. \(a^2 b = c^2 d\)
• B. \(a^2 c = b^2 d\)
• C. \(a^2 d = c^2 b\)
• D. \(d^2 b = c^2 a\)

Hint:

Write the roots as mk and k using the given ratio, express b/a^2 in terms of the ratio alone, and do the same for d/c^2; the two expressions must then be equal.

Answer 1

The Correct Option is C

Step 1: Write the roots using the quadratic formula.
For $x^2 - 2ax + b = 0$, the roots are $a + \sqrt{a^2 - b}$ and $a - \sqrt{a^2 - b}$. Call the ratio of the roots $\lambda$, taking the larger root over the smaller one:
\[ \lambda = \frac{a + \sqrt{a^2-b}}{a - \sqrt{a^2-b}} \]

Step 2: Apply componendo and dividendo.
This is a standard trick for ratios written as (sum)/(difference): if $\dfrac{p+q}{p-q} = \lambda$, then $\dfrac{\lambda+1}{\lambda-1} = \dfrac{p}{q}$. Applying it here with $p = a$ and $q = \sqrt{a^2-b}$:
\[ \frac{\lambda+1}{\lambda-1} = \frac{a}{\sqrt{a^2-b}} \]
Squaring both sides removes the square root:
\[ \frac{(\lambda+1)^2}{(\lambda-1)^2} = \frac{a^2}{a^2-b} \]

Step 3: Do the same for the second equation.
Since the roots of $x^2 - 2cx + d = 0$ share the same ratio $\lambda$, the identical working gives:
\[ \frac{(\lambda+1)^2}{(\lambda-1)^2} = \frac{c^2}{c^2-d} \]

Step 4: Equate the two right-hand sides.
Both equal the same left-hand expression $\dfrac{(\lambda+1)^2}{(\lambda-1)^2}$, so:
\[ \frac{a^2}{a^2-b} = \frac{c^2}{c^2-d} \]
Cross-multiply:
\[ a^2 (c^2 - d) = c^2 (a^2 - b) \]
\[ a^2 c^2 - a^2 d = a^2 c^2 - c^2 b \]
The $a^2 c^2$ term cancels from both sides, leaving:
\[ -a^2 d = -c^2 b \quad \Rightarrow \quad a^2 d = c^2 b \]

Final Answer:
This matches the same relation found through the sum-and-product method, confirming option (C) is correct.
\[ \boxed{a^2 d = c^2 b} \]