Answered

Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Connect with professionals on our platform to receive accurate answers to your questions quickly and efficiently. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

16x²-81=0 solve the following quadratic equation

Sagot :

deathstrike001
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 hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.