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 $3x^2 -12x + 11 = 0$

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

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

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

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

How do you solve 3x^2-12x+11=03x^2-12x+11=0 using the quadratic function?