How do you solve #2/3 x^2 + 1/4 x= 3#?

1 Answer
May 17, 2018

Answer:

See a solution process below:

Explanation:

First, multiply each side of the equation by #color(red)(12)# to eliminate the fractions:

#color(red)(12)(2/3x^2 + 1/4x) = color(red)(12) xx 3#

#(color(red)(12) xx 2/3x^2) + (color(red)(12) xx 1/4x) = 36#

#8x^2 + 3x = 36#

Next, put the equation in standard quadratic form:

#8x^2 + 3x - color(red)(36) = 36 - color(red)(36)#

#8x^2 + 3x - 36 = 0#

We can nowuse the quadratic equation to solve this problem:

The quadratic formula states:

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)(8)# for #color(red)(a)#

#color(blue)(3)# for #color(blue)(b)#

#color(green)(-36)# for #color(green)(c)# gives:

#x = (-color(blue)(3) +- sqrt(color(blue)(3)^2 - (4 * color(red)(8) * color(green)(-36))))/(2 * color(red)(8))#

#x = (-color(blue)(3) +- sqrt(9 - (-1152)))/16#

#x = (-color(blue)(3) +- sqrt(9 + 1152))/16#

#x = (-color(blue)(3) +- sqrt(1161))/16#

#x = (-color(blue)(3) +- sqrt(9 * 129))/16#

#x = (-color(blue)(3) +- sqrt(9)sqrt(129))/16#

#x = (-color(blue)(3) +- 3sqrt(129))/16#

#x = (3(-1 +- sqrt(129)))/16#

Or

#x = 3/16(-1 +- sqrt(129))#