How do you solve #4x^2-64=0# using the quadratic formula?

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Answer 1 · 2017-08-19T15:03:50

See a solution process below:

We can rewrite the equation as:

#4x^2 + 0x - 64 = 0#

For #color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0#, the values of #x# which are the solutions to the equation are given by:

#x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))#

Substituting:

#color(red)(4)# for #color(red)(a)#

#color(blue)(0)# for #color(blue)(b)#

#color(green)(-64)# for #color(green)(c)# gives:

#x = (-color(blue)(0) +- sqrt(color(blue)(0)^2 - (4 * color(red)(4) * color(green)(-64))))/(2 * color(red)(4))#

#x = +- sqrt(0 - (-1024))/8#

#x = +- sqrt(1024)/8#

#x = +- 32/8#

#x = +- 4#