Can you solve for #x# in #ax^2 - bx + c = 0# (if a,b,c are constants)?

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Answer 1 · 2018-02-16T14:02:08

See a solution process below:

#ax*x - bx + c = 0# can be rewritten as:

#ax^2 - bx + c = 0#

We can use the quadratic equation to solve this problem:

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

#x = (-color(blue)(m) +- sqrt(color(blue)(m)^2 - (4color(red)(l)color(green)(n))))/(2 * color(red)(l))#

Substituting:

#color(red)(a)# for #color(red)(l)#

#color(blue)(-b)# for #color(blue)(m)#

#color(green)(c)# for #color(green)(n)# gives:

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

Or

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