Question

Find the zeroes of 49x2-64

Answer 1

Answer:

-8/7 and 8/7

Step-by-step explanation:

49x2-64 = (7x)^2-8^2

we know that a^2-b^2 = (a+b)(a-b)

therefore,

              (7x)^2-8^2 = (7x+8)(7x-8)

Hence,

           7x+8 = 0

                 7x = -8

                  x = -8/7

             and

          7x-8 = 0

                7x = 8

                  x = 8/7

Answer 2

Hi!! Here's your answer...

$\mapsto\ 49x^2-64 \\ \\ \mapsto\ (7x)^2-(8)^2 \\ \\ \mapsto\ (7x+8)(7x-8) \\ \\ \\ \therefore\ \boxed{x=\bold{\pm \frac{8}{7}}}$

Thank you...