How do you solve #81x^2=9# using the quadratic formula?

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Answer 1 · 2017-07-02T19:46:18

See a solution process below:

First, we must put the equation in standard quadratic form which is:

#ax^2 + bx + c = 0#

#81x^2 = 9#

#(81x^2)/color(red)(9) = 9/color(red)(9)#

#9x^2 = 1#

#9x^2 - color(red)(1) = 1 - color(red)(1)#

#9x^2 - 1 = 0#

Because there is no #x# term in the quadratic the coefficient for the #x# term is #0#:

#9x^2 + 0x - 1 = 0#

For #ax^2 + bx + c = 0#, the values of #x# which are the solutions to the equation are given by:

#x = (-b +- sqrt(b^2 - 4ac))/(2a)#

Substituting #9# for #a#; #0# for #b# and #-1# for #c# gives:

#x = (-0 +- sqrt(0^2 - (4 * 9 * -1)))/(2 * 9)#

#x = +- sqrt(0 - (-36))/18#

#x = +- sqrt(36)/18#

#x = +- 6/18#

#x = +- 1/3#