use the quadratic formula to find the roots: 3x^2+6x=-2. answer in radical form

Answer 1

3x^2 +6x = -2

divide by 3

x^2 +2x = -2/3

complete the square

x^2 +2x +1 = -2/3 + 1

(x+1)^2 = 1/3

take square roots of both sides

x+1 = + or - sqr(1/3)

x = -1 + 1/sqr3 or -1 - 1/sqr3

= -1 + (sqr3)/3

but you wanted to use the quadratic formula, so

rewrite the given equation in y = ax^2 + bx + c form

where a= 3, b =6 and c= 2

then plug those values into the quadratic formula

x= -b/2a + or - (1/2a)sqr(b^2 -4ac)

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

x= -1 + or - (1/6)sqr(36 -24)

x = -1 + or - (1/6)sqr(12)

x = -1 + or - (1/6)(2)sqr3

x = -1 + or - (2/6)sqr3

x= -1 + (sqr3)/3 and -1 - (sqr3)/3 = the two roots.

= about -1 + or minus 1.732/3 = -1+ 0.577 = -1.577 or -0.423 rounded off to 3 decimal places

the two roots are a conjugate pair

however you solve the equation, you should get the same answer. It's a good check, to try it two different ways. When you get the same answer, you're more confident it's the correct answer.

quadratic equations have potentially two roots, either 2 real roots, 1 real root with multiplicity 2, or 2 imaginary roots. By Descartes' Method there are zero positive real roots, and either 2 or 0 negative roots.

Graphically, the roots are x intercepts of an upward opening parabola

with axis of symmetry x= -1 and vertex = (-1, -1)= minimum point

sometimes it helps to sketch the graph. The two roots or x intercepts will be equidistant to the axis of symmetry, x=-1. If there are no x intercepts, if the upward opening parabola were completely above the x axis, then the roots are imaginary. If the vertex had been on the x axis, then there would have been one root, with multiplicity two.