How do you solve #3x^2+9x-6=0#?

1 Answer
Apr 8, 2017

The answer is
#x=-(3+sqrt(17))/2# and #x=-(3-sqrt(17))/2#

Explanation:

The above equation is what we called Quadratic Equation.There are 2 types of quadratic equation

  1. Pure Quadratic equation
    Example: #x^2-a=0#
  2. Affected Quadratic Equation
    Example:#ax^2+bx+c=0#

So,the stated equation is a affected quadratic equation
General from of a affected quadratic equation is

#color(red)(ax^2+bx=c=0)#........#(i)#

Where a(#!=0#) and b are the coefficients of #x^2#and #x# respectively and c is constant term.

As #a!=0#, From #(i)# we get,

#4a^2x^2+4abx+4ac=0[(i)xx4a]#
#⇒4a^2x^2+4abx+b^2-b^2+4ac=0#
#⇒(2ax+b)^2=b^2-4ac#
#⇒2ax+b=+-sqrt(b^2-4ac)#
#2ax=-b+-sqrt(b^2-4ac)#
#color(DEEPSKYBLUE)(x=(-b+-sqrt(b^2-4ac))/(2a))#

So we get two roots which are denoted by α and β
#color(DARKSALMON)(α=(-b+sqrt(b^2-4ac))/(2a))# and #color(INDIANRED)(β=(-b+sqrt(b^2-4ac))/(2a))#

So using this formula,we get,
#x=(-9+-sqrt(9^2-4*3*-6))/(2*3)#
#⇒x=(-9+-sqrt(81+72))/6#
#⇒x=(-9+-sqrt(153))/6#
#⇒x=(-9+-3sqrt17)/6#
#⇒x=-cancel(3)*(3+-sqrt17)/(cancel(6)2)#
#color(SIENNA)(x=-(3+sqrt17)/2)# and #color(SEAGREEN)(x=-(3-sqrt(17))/2)#