Question

Solve: 16x2 – 8a2x + (a4 – b4)= 0 for x​

Answer 1

Given: The equation 16x^2 – 8a^2x + (a^4 – b^4)= 0

To find: The value for x​ .

Solution:

• Now we have given the equation as:

                   16x^2 – 8a^2x + (a^4 – b^4) = 0

• For roots, we have:

                   x = -b ± √D / 2a

                   x = 8a^2 ± √(64a^4 - 4(16)(a^4 – b^4)) / 2(16)

                   x = 8a^2 ± √(64a^4 - 64(a^4 – b^4)) / 32

• Taking 64 common as root, we get:

                   x = 8a^2 ± 8√a^4 - (a^4 – b^4) / 32

• Cancelling multiples of 8, we get:

                   x = a^2 ± √a^4 - a^4 + b^4 / 4

                   x = a^2 ± √b^4 / 4

                   x = a^2 ± b^2 / 4

Answer:

             So the value of x is a^2 ± b^2 / 4

Answer 2

Answer:

x=(a²+b²)/4 or (a²-b²)/4

Step-by-step explanation:

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