Question

Solve : 16xsquare-8asquarex+(a^4-b^4)=0 for x

Answer 1

I think the answer is sec c

Answer 2

Answer:

x=\frac{1}{4}(a^2+b^2),\frac{1}{4}(a^2-b^2)

Step-by-step explanation:

The given equation is 16x^2-8a^2x+(a^4-b^4)=0

We can rewrite this equation as

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

We can write  16x^2-8a^2x+a^4=(a^2-4x)^2

Thus, the equation becomes

(a^2-4x)^2-b^4=0\\\\(a^2-4x)^2-(b^2)^2=0

Apply the formula a^2-b^2=(a+b)(a-b)

(a^2-4x+b^2)(a^2-4x-b^2)=0

Apply the zero product property

(a^2-4x+b^2)=0, (a^2-4x-b^2)=0

Solve the equations for x

x=\frac{1}{4}(a^2+b^2),x=\frac{1}{4}(a^2-b^2)