Question

4x²-4a²x + (a¹-b¹)=0​

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Answer 1

Answer:

4x²-4a²x+(a⁴-b⁴)=0

or, 4x²-2{(a²+b²)+(a²-b²)}x+(a²+b²)(a²-b²)=0

or, 4x²-2(a²+b²)x-2(a²-b²)x+(a²+b²)(a²-b²)=0

or, 2x{2x-(a²+b²)}-(a²-b²){2x-(a²+b²)}=0

or, {2x-(a²+b²)}{2x-(a²-b²)}=0

Either, 2x-(a²+b²)=0

or, 2x=a²+b²

or, x=(a²+b²)/2

Or, 2x-(a²-b²)=0

or, 2x=a²-b²

or, x=(a²-b²)/2

∴, x=(a²+b²)/2, (a²-b²)/2 Ans

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

4x²-4a²+(a-b)=0 Let

a=4 b=-4 c-a-b

By applying formula

D=b²-4ac

(-4)²- 4x4×(a-b)

16-16(a-b)

16[1-(a-b)]

Solve for root

x=-b+- √D/2a

-(-4)+-√16(1-(a-b)/2x4

4-16√(1-a+b)/8