How do you solve 49x^2 - 64 = 0?

Jun 27, 2016

$x = \frac{8}{7} \mathmr{and} x = - \frac{8}{7}$

Explanation:

$49 {x}^{2} - 64 = 0 \mathmr{and} {\left(7 x\right)}^{2} - {\left(8\right)}^{2} = 0 \mathmr{and} \left(7 x + 8\right) \left(7 x - 8\right) = 0 \therefore \left(7 x - 8\right) = 0 \mathmr{and} \left(7 x + 8\right) = 0 \therefore x = \frac{8}{7} \mathmr{and} x = - \frac{8}{7}$[Ans]

Jun 27, 2016

$x = = \frac{8}{7} \mathmr{and} x = - \frac{8}{7}$

Explanation:

Usually we make a quadratic equal to 0 to be able to solve it.
This is an exception, where we can use the normal method of x on one side and the numbers on the other.

$49 {x}^{2} = 64$

${x}^{2} = \frac{64}{49} \text{ find the square root of both sides}$

$x = \pm \sqrt{\frac{64}{49}}$

$x = = \frac{8}{7} \mathmr{and} x = - \frac{8}{7}$

However, we can also use the factorising method:

$49 {x}^{2} - 64 = 0 \text{ the difference of two squares}$

$\left(7 x + 8\right) \left(7 x - 8\right) = 0$

$x = - \frac{8}{7} \mathmr{and} x = \frac{8}{7}$