Question

Solve: 16x2 – 8a2x + (a4 – b
4
) = 0 for x​

Answer 1

Answer:

16×2=32-8a2x+4-b =as8a-a4=4a -b so the answer is 32-4a2x-b

Answer 2

Answer:

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

Explanation: