Ok Since this cant be solved by factoring?

Question

x^2+3x+12=0
how do you solve it? cause i thought that you solved these by factoring.

Answer 1

There are a couple of different ways. You could use the quadratic formula, or completing the square.

The quadratic formula is
x = [ -b ± √(b²-4ac) ] / 2a
where a, b and c are the coefficients when the equation is put into the form of ax² + bx + c = 0. In this case we have

x = [ -3 ± √(9 - 4*1*12) ] / 2
x = [ -3 ± √(-39) ] / 2
This involves taking the square root of a negative number, so in this case there are NO REAL SOLUTIONS.

Here's the solution via completing the square:
x² + 3x + 12 = 0
x² + 3x = -12
x² + 3x + 9/4 = -12 + 9/4
(x + 3/2)(x + 3/2) = -48/4 + 9/4
(x + 3/2)² = -39/4

Here you have a square root that equals a negative number, so as shown before, there's no solution.

Answer 2

This is a quadratic equation (highest degree is 2) and only SOME quadratics can be solved by factoring. Most actually can't (although your algebra book probably misleads you!). In cases where you can't factor, you have a couple other choices: quadratic formula or completing the square. There are also some calculators that will solve equations for you, either by graphing or other methods.

The quadratic formula says that if an equation is in the form
ax^2 + bx + c = 0, then
x= (-b +- sqrt (b^2-4ac)) / (2a)

here you have:

x = (-3 +- sqrt (9-48)) / 2
x = (-3 +- sqrt (-39)) / 2
x = (-3 +- i sqrt 39) / 2

Not only was your equation prime, but it has imaginary solutions.

Answer 3

you will possibly desire to learn how to remedy those basic problems with out calculator. or you in no way learn the undemanding regulations. in fact, i might in no way think of of a calculator for those. -11x ? 2 = -11x + ? ? = –2 for sure x2 + x ? 2 = 0 assuming it truly is x² + x ? 2 = 0 please use prevalent terminology. use x² or x^2 (x + 2)(x – a million) = 0 x = –2, +a million -5x ? 10y = -7 2x + 2y = 2 multiply 2d by ability of five and upload 10x + 10y = 10 –5x ? 10y = –7 5x = 3 x = 3/5 2x + 2y = 2 x + y = a million (3/5) + y = a million y = 2/5 .

Answer 4

Your equation can be solve by completeing the square or by the quadratic formula.
X = (-B+-(B^2-4AC)^.5)/(2A)

X = (-3+-(3^2 - 4*1*12)^.5)/(2*1)
The solution will involve imaginary numbers

Answer 5

You have a formula to solve these type of equation

factors is given by

x1= -b+(b^2-4ac)^(1/2)
x2= -b-(b^2-4ac)^(1/2)

where
a is term with x^2
b is term with x
c is constant

thus equation is given by
(x-x1)(x-x2)

Answer 6

Ha ha I just learned this today. you do: -b +or- the square root of b^-4ac over 2a. Just in case you were wondering, a wouls equal 1, b would equal 3 and c would equal 12.

Answer 7

Use the quadratic formula.

x = (-b ± sqrt(b^2- 4ac))/(2a)

x = (-3 ± sqrt(3^2 - 4(1)(12)))/(2(1))
x = (-3 ± sqrt(9 - 48))/2
x = (-3 ± sqrt(-39))/2
x = (-3 ± isqrt(39))/2

ANS : (1/2)(-3 ± sqrt(39))

Answer 8

it is of the form ax^2 + bx + c

x = -b + (sqr. root of (b^2 - 4ac))/2a

or

x = -b - (sqr. root of (b^2 - 4ac))/2a
use this formula!

so, it's -3 + (sqr. root of (9 - 48))/2 or -3 - (sqr. root of (9 - 48))/2

Answer 9

-b(+or-) b^2 square root b^2-4(a)(c) over 2(a)

Answer 10

(-b +/- squareroot of b*2-4ac) DIVIDED BY 2a
its quadrastic! :D
ax*2 + bx +c