Question

(a²+b²)x² - 2(ac+bd)x + c² + d² = 0
If the following equation has equal roots, then
a) ab = cd b) ad = bc c) ad = √bc d) ab = √cd

please answer

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Answer 1

Answer:

Option b) ad = bc

Step-by-step explanation:

For equal roots of Ax²+Bx+C=0, B²-4AC=0

Here, A=(a²+b²), B= - 2(ac+bd), C=(c²+d²)

∴ B²-4AC=[ - 2(ac+bd)]² - 4(a²+b²)(c²+d²)

Since B²-4AC=0, B² = 4AC

∴ [ - 2(ac+bd)]² = 4(a²+b²)(c²+d²)

4[(a²c²+b²d²+2abcd)] = 4[a²(c²+d²)+b²(c²+d²)] = 4[a²c²+a²d²+b²c²+b²d²)]

∴ a²c²+b²d²+2abcd = a²c²+a²d²+b²c²+b²d²

(cancelling similar terms on both sides)

∴ 2abcd = a²d²+b²c²

or (ad)²+(bc)²=2abcd

(ad)²+(bc)²-2abcd = 0

∴ (ad-bc)²=0

or ad-bc=0

∴ ad = bc

Answer 2

Answer:

(a^2 + b^2)x^2 - 2(ac + bd)x + (c^2 + d^2) = 0 has equal roots.Therefore, discriminant = 0 Thus [2(ac + bd)]^2 + 4(a^2 + b^2) (c^2 + d^2) a^2c^2 + ...