Question

27. 3a232 + 8abx + 4b2 = 0, a ≠ 0​

Answer 1

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(i) 3a2x2 + 8abx + 4b2 = 0

⇒ 3a2x2 + 6abx + 2abx + 4b2 = 0

⇒ 3ax(ax + 2b) + 2b(ax + 2b) = 0

⇒ (3ax + 2b) (ax + 2b) = 0

⇒ x = - 2b/3a or - 2b/a

(ii) (x – a/b)2 = a2/b2

⇒ x2 + a2/b2 – 2ax/b = a2/b2

⇒ x2 – 2ax/b = 0

⇒ x (x – 2a/b) = 0

⇒ x = 0 or x – 2a/b = 0

⇒ x = 0 or x = 2a/b.

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Answer 2

Answer:

Step-by-step explanation:

Compare 3a²x²+8abx+4b²=0 with Ax²+Bx+C=0 , we get

A=3a², B=8ab , C = 4b²

Discreminant (D)=B²-4AC

= (8ab)²-4×(3a²)(4b²)

= 64a²b²-48a²b²

= 16a²b²--(1)

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