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How do you solve #6x²-81=0#?

Answer 1

Ethan Adkins

We can manipulate the expression by isolating #x#. Alternatively, we can proceed to use Bhaskara.

Isolating #x#

#6x^2-81=0# #x^2=81/6# #x=sqrt(81/6)# #x=+-9/sqrt(6)# Rationalizing: #x=+-(3sqrt(6))/6#

Using Bhaskara, where #a=6#, #b=0# and #c=-81#

#(-0+-sqrt(0-4(6)(-81)))/12# #(+-sqrt(1944))/12# #(+-sqrt(2^3*3^5))/12# #(+-18sqrt(6))/12#

#+-(3sqrt(6))/2#

P.s.: there are other forms, but these seem so simple they appeal to me!

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Answer 2

Abigail Allen

To solve the equation 6x² - 81 = 0, you can use the method of factoring or the quadratic formula. Factoring the quadratic equation, you get (2x + 9)(3x - 9) = 0. Then, set each factor equal to zero and solve for x. So, 2x + 9 = 0 gives x = -9/2, and 3x - 9 = 0 gives x = 3. Therefore, the solutions are x = -9/2 and x = 3.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.