Question

16x²-81=0 solve the following quadratic equation

Answer 1

$16x^2-81=0\\\\(4x)^2-9^2=0\\\\(4x-9)(4x+9)=0\iff4x-9=0\ \vee\ 4x+9=0\\\\4x=9\ \vee\ 4x=-9\\\\x=\frac{9}{4}\ \vee\ x=-\frac{9}{4}$


$or\\\\16x^2-81=0\\\\16x^2=81\ \ \ \ /:16\\\\x^2=\frac{81}{16}\\\\x=\pm\sqrt\frac{81}{16}\\\\x=-\frac{\sqrt{81}}{\sqrt{16}}\ \vee\ x=\frac{\sqrt{81}}{\sqrt{16}}\\\\x=-\frac{9}{4}\ \vee\ x=\frac{9}{4}$

Answer 2

16x^2-81=0
(4x-9)(4x+9)
x=9/4 or x=-9/4