How do you solve #3x^2-12x+11=0# using the quadratic function?

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Answer 1 · 2018-06-27T06:55:34

#x=(6+-sqrt3)/3#

The quadratic equation is

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

for a quadratic taking the form

#ax^2+bx+c=0#

So for this equation it is

#x=(-(-12)+-sqrt((-12)^2-4*3*11))/(2*3)#

#x=(12+-sqrt(144-132))/(6)#

#x=(12+-sqrt(4*3))/6#

#x=(6+-sqrt3)/3#

Answer 2 · 2018-07-12T02:55:22

#color(blue)(x = 2 + 1/sqrt3, 2 - 1/sqrt3= 2.5774, 1.4226#

Quadratic formula to find the roots #x = (-b +- sqrt D) / (2a)# where D is the discrimination to decide whether the roots are real or imaginary.

#D = b^2 - 4 a c#

Given equation is #3x^2 -12x + 11 = 0#

#a = 3, b = -12, c = 11, D = -12^2 - (4 * 3 * 11) = 12#

#x = (12 +- sqrt 12) / 6#

#x = 2 +- (1/sqrt3)#

#color(blue)(x = 2 + 1/sqrt3, 2 - 1/sqrt3= 2.5774, 1.4226#