If the roots of the equation $(a^2 + b^2)x^2 - 2(ac + bd)x + (c^2 + d^2) = 0$ are equal, prove that $\dfrac{a}{b} = \dfrac{c}{d}$.
Step-by-Step Solution
Key Concept: $D = 4(ac + bd)^2 - 4(a^2 + b^2)(c^2 + d^2) = 0 \Rightarrow (a^2 c^2 + 2abcd + b^2 d^2) - (a^2 c^2 + a^2 d^2 + b^2 c^2 + b^2 d^2) = 0 \Rightarrow 2abcd - a^2 d^2 - b^2 c^2 = 0 \Rightarrow -(ad - bc)^2 = 0 \Rightarrow ad = bc \Rightarrow \dfrac{a}{b} = \dfrac{c}{d}$.
$D = 4[(ac + bd)^2 - (a^2 + b^2)(c^2 + d^2)] = 0$. [1.5 Marks]
Expand: $(a^2 c^2 + 2abcd + b^2 d^2) - (a^2 c^2 + a^2 d^2 + b^2 c^2 + b^2 d^2) = 0$. [1.5 Marks]
$2abcd - a^2 d^2 - b^2 c^2 = 0 \Rightarrow -(ad - bc)^2 = 0 \Rightarrow ad - bc = 0 \Rightarrow ad = bc \Rightarrow \dfrac{a}{b} = \dfrac{c}{d}$. Proved! [2.0 Marks]
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🎯 Official CBSE Marking Scheme:
Setting up $D = 0$: 1.5 Marks
Expanding terms and simplifying: 1.5 Marks
Proving $(ad - bc)^2 = 0 \Rightarrow a/b = c/d$: 2.0 Marks
Correct Answer: