27. 3a232 + 8abx + 4b2 = 0, a ≠ 0
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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.
Answered by
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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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