Question

16x²-81=0 solve the following quadratic equation

Answer 1

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

Answer 2

`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=(9)/(4)\ \vee\ x=-(9)/(4)`



`or\\\\16x^2-81=0\\\\16x^2=81\ \ \ \ /:16\\\\x^2=(81)/(16)\\\\x=\pm\sqrt(81)/(16)\\\\x=-(√(81))/(√(16))\ \vee\ x=(√(81))/(√(16))\\\\x=-(9)/(4)\ \vee\ x=(9)/(4)`