Q.If the roots of the quadratic equation \((a^2 + b^2)x^2 - 2(ac + bd)x + (c^2 + a^2) = 0\) are equal, prove that \(\cfrac{a}{b} = \cfrac{c}{d}\).

\((a^2 + b^2) x^2 + 2(ac + bd)x\) \(+(c^2 + d^2) = 0 \)

Since the roots of the quadratic equation are equal, the discriminant must be \(0\).

\(∓ {2(ac + bd)}^2 - 4(a^2 + b^2)(c^2 + d^2) = 0\)
Or, \(4(a^2 c^2 + 2abcd + b^2 d^2)\)

\(- 4(a^2 c^2 + a^2 d^2 + b^2 c^2 + b^2 d^2) = 0\)



Or, \(4a^2 c^2 + 8abcd + 4b^2 d^2 - 4a^2 c^2 - 4a^2 d^2\)

\(- 4b^2 c^2 - 4b^2 d^2 = 0\)



Or, \(-4a^2 d^2 + 8abcd - 4b^2 c^2 = 0\)
Or, \(a^2 d^2 - 2abcd + b^2 c^2 = 0\)
Or, \((ad)^2 - 2.ad.bc + (bc)^2 = 0\)
Or, \((ad - bc)^2 = 0\)
Or, \(ad - bc = 0\)
Or, \(ad = bc\)
Or, \(\cfrac{a}{b} = \cfrac{c}{d}\) (Proved).
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