Mason M.

asked • 11d# How do i find the missing root

3x²+bx+2

## 2 Answers By Expert Tutors

completing square and solving for x is same as using quadratic formula

set equation to zero

divide by 3

have x^{2}+bx/3+2/3=0

same as

(x+b/6)^{2}-b^{2}/36+2/3=0

x+b/6=+/-√(b^{2}/36-2/3)

x=-b/6+/-√(b^{2}/36-2/3)

note you have two roots that make equation true (equal zero)

so 3(x-[-b/6+√(b^{2}/36-2/3)(x-(-b/6-√(b^{2}/36-2/3)=3x^{2}+b+2

which equals zero when x is equal to either of the two roots.

two roots are equal in this case if b^{2}/36-2/3=0

this is called also a root that is duplicated

if that is required then b=+/-2√6

that is to say f(x)=3x^{2-}(2√6)x+2=3(x-(2√6)/6)^{2},,,,recall b/6 is in factored form of equation

equation has one root x=+(√6)/3 corresponding to b=-2√6 (see (2√6)/6=(√6)/3)

or equation f(x)=3x^{2}+(2√6)x+2=3(x-(-2√6)/6)^{2}

equation has one root x=-(√6)/3 corresponding to b=+2√6

3x^2 + bx + 2 = 0

use the quadratic formula, x =-b/2a +/- (1/2a)sqr(b^2-4ac), a =3, c=2

x = -b/6 + or - (1/6)sqr(b^2 -4(3)(2)

x = (-b +/-sqr(b^2-24)/6

two real roots if b^2>24, two imaginary roots if b^2<24, one real root if b^2 =24 or you could, if you really want to, say there's 2 roots, but they repeat or that the two roots are the same identical number.

if (-b +sqr(b^2-24))/6 is one root, then the other root is

(-b-sqr(b^2-24))/6

or solve for b:

bx = -3x^2 -2

b = -3x -2/x

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Mason M.

and whats the answer11d