Answered

Experience the convenience of getting your questions answered at IDNLearn.com. Join our knowledgeable community and get detailed, reliable answers to all your questions.

16x²-81=0 solve the following quadratic equation

Sagot :

Deathstrike001idn
16x^2-81=0
(4x-9)(4x+9)
x=9/4 or x=-9/4
[tex]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}[/tex]


[tex]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}[/tex]
We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Trust IDNLearn.com for all your queries. We appreciate your visit and hope to assist you again soon.