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Hello people ~
factorise: x^2 + 3 √3 x + 6​

User Tws
by
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2 Answers

9 votes
9 votes

Using the quadratic formula,

∆=b²-4ac

∆=(3√3)²-4(1)(6)

∆=(9×3)-4(6)

∆=27-24

∆=3


x1 = ( - b - √(3) )/(2a) = ( - 3 √(3) - √(3) )/(2) = \frac{ - 4 \sqrt[]{3} }{2}


x1 = - 2 √(3)

Similarly,


x 2 = ( - b + √(3) )/(2a) = \frac{ - 3 √(3) + \sqrt[]{3} }{2} = ( - 2 √(3) )/(2)


x2 = - √(3)

Method 2:

x²-Sx+P=0

where S is the sum of the roots & P is the product of the roots.

x¹+x²= -3√3

x¹x²=6

Solving the system you get the same answers.

Your factored equation can be written in the form of:

(x-x¹)(x-x²)

(x+2√3)(x+√3)

User Pinckney
by
2.6k points
23 votes
23 votes


\\ \rm\rightarrowtail x^2+3√(3)x+6


\\ \rm\rightarrowtail x^2+√(3)x+2√(3)x+6


\\ \rm\rightarrowtail x(x+√(3))+2√(3)(x+√(3))


\\ \rm\rightarrowtail (x+2)(x+2√(3))

User Emilio M Bumachar
by
2.3k points