find roots 9x2-6b2x-a4-b4
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Answered by
59
9x2 – 6ax + (a – b)(a + b) = 0
9x2 – 3x(a – b) – 3x(a + b) + (a – b)(a + b) = 0
3x[3x – (a – b)] – (a + b) [3x – (a – b)] = 0
[3x – (a – b)][3x – (a + b)] = 0
[3x – (a – b)] = 0 or [3x – (a + b)] = 0
3x = (a – b) or 3x = (a + b)
∴ x = (a – b)/3 or x = (a + b)/3
9x2 – 3x(a – b) – 3x(a + b) + (a – b)(a + b) = 0
3x[3x – (a – b)] – (a + b) [3x – (a – b)] = 0
[3x – (a – b)][3x – (a + b)] = 0
[3x – (a – b)] = 0 or [3x – (a + b)] = 0
3x = (a – b) or 3x = (a + b)
∴ x = (a – b)/3 or x = (a + b)/3
kumarrohit321:
Thanks
But bro in the question it is a4-b4 not a2-b2
oh
so u can comvert it into (a²-b²)+(a²+b²)
ya see a²-b²=(a-b)+(a+b)
Answered by
46
Answer:
Step-by-step explanation:
9x²-6b²x-(a^4-b^4)=0
9x²-3(2b²)x-(a^4-b^4)=0
9x²-3(b²-a²+b²+a²)x+(b^4-a^4)=0
9x²-3(b²-a²)x-3(b²+a²)x+(b²+a²)(b²-a²)=0
3x(3x-b²+a²)-(b²+a²)(3x-b²+a²)=0
(3x-b²-a²)(3x-b²+a²)=0
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