Algebra Question 2x^2+4x=-3 ?

You can either put them in the form of:

ax² + bx + c = 0

and factor if you can, or use quadratic equation, or in the form of:

x² + bx = c

and complete the square.

I'll complete the square on the first one:

2x² + 4x = -3

divide all by 2:

x² + 2x = -3/2

Half the coefficient of the "x" and square it, then add that to both sides. In this case, it's 1:

x² + 2x + 1 = -3/2 + 1

simplify:

(x + 1)² = -1/2
x + 1 = ± √(-1/2)
x = -1 ±√(-1/2)

Now simplify to get the fraction and negative out of the radical:

x = -1 ±√(-1/2)
x = -1 ± i/√(2)
x = -1 ± i√(2)/2


For #2: I'll factor:

2x² - 32x = 0
2x(x - 16) = 0

Since the product of anything and 0 is 0, I'll set both of these to 0 and solve for x:

2x = 0
x = 0

x - 16 = 0
x = 16

x = 0 and 16.


#3, put it into a standard form first:

2x(x - 4) = 3(1 - x)
2x² - 8x = 3 - 3x
2x² - 5x - 3 = 0

I'll use quadratic formula here:

x = [-b ± √(b² - 4ac)] / 2a
x = [-(-5) ± √((-5)² - 4(2)(-3))] / 2(2)
x = [5 ± √(25 + 24)] / 4
x = [5 ± √(49)] / 4
x = [5 ± 7] / 4
x = -2/4 and 12/4
x = -1/2 and 3

Since that came out as rational numbers, this could have been factored, but as I said, either method would work.

2x^2-32x=0

2x^2=32x

2x=32

---x=16.---

2x(x-4)=3(1-x)

2x^2-8x=3-3x

2x^2-5x-3x=0

[5 (+or-) sqrt(25+24)]/4

=> [5(+or-)7]/4
=3 and (-0.5)


check it out.. http://bux.to/?r=uydarp

8x^6+3x^5+2x^4 8x^3-28x^2+4x-6x^2+21x-3= 8x^3-34x^2+25x-3 x^2+4x -3 2x^2+x+6 -------------------------- 6x^2 +24x -18 x^3 +4x^2 -3x 2x^4+8x^3 -6x^2 --------------------------------------... 2x^4+9x^3 +4x^2 +21x -18

hi friends, before you look in the answer i suggest you watch these videos, it's very helpful.here is the link:
http://mathorhealth.blogspot.com/2008/07/quadratic-eq-2.html
the answer is :
1). 2x^2 + 4x = -3, 2x^2 + 4x + 3= 0
because we can not do by factorization method
we'll do it by completing the square method
1st. divided all by 2, u'll get...
x^2 + 2x = - 3/2
2nd. add (b/2)^2 which is 1.
x^2 + 2x + 1 = -1/2
( x + 1)^2 = -1/2
take square root
x + 1 = + - square root of -1/2
x = + - square root of -1/2 - 1
these are the complex roots.
but if the domain = real numbers , the answer is no real roots.
for no 2.
2x^2 - 32 x = 0
x^2 - 16x = 0
x( x -16) = 0
x =0 or x= 16

no 3.
2x^2 - 8x = 3 - 3x
2x^2 - 8x + 3x -3 = 0
2x^2 - 5x - 3 = 0
(2x+1)(x-3) = 0
2x + 1 = 0 or x-3=0
x = -1/2 or x= 3

ciaoo....

arrange the equations this way ax^2 +bX + C = 0

2 X^2 +4X+ 3 = 0

using quadratic formula X= -4 +/- {SQRT( 16-4(2)(3) }/4
X = -4 +/- SQRT ( -8)/4 = -4 +/- 2i SQRT 2/4 =( -2+/- iSQRT 2)/2

2X^2-32 =0 , divide each term by 2 X^2 -16 =0 X^2 =16 X =+/- 4

expand the terms 2X(X-4) = 2 X^2 -8X, 3(1-x) = 3 -3X

then simplify 2 X^2-8X = 3 -3X add 3x to each side and subtract 3 from each side 2X^2-5X-3 = 0 this actors into (2X+1))X-3)

set each equal to zero 2X+1 = 0, X = -1/2 X-3 = 0 X = 3

1)
2x^2 + 4x = -3
2x^2 + 4x + 3 = 0
x = [-b ±√(b^2 - 4ac)]/2a

a = 2
b = 4
c = 3

x = [-4 ±√(16 - 24)]/4
x = [-4 ±√-8]/4 (imaginary number)
(no real roots)

= = = = = = = =

2)
2x^2 - 32x = 0
(2x^2 - 32x)/2 = 0/2
x^2 - 16x = 0
x(x - 16) = 0

x = 0

x - 16 = 0
x = 16

∴ x = 0 , 16

= = = = = = = =

3)
2x(x - 4) = 3(1 - x)
2x*x - 2x*4 = 3*1 - 3*x
2x^2 - 8x = 3 - 3x
2x^2 - 8x + 3x - 3 = 0
2x^2 - 5x - 3 = 0
2x^2 + x - 6x - 3 = 0
(2x^2 + x) - (6x + 3) = 0
x(2x + 1) - 3(2x + 1) = 0
(2x + 1)(x - 3) = 0

2x + 1 = 0
2x = -1
x = -1/2 (-0.5)

x - 3 = 0
x = 3

∴ x = -1/2 (-0.5) , 3

Question 1
2x² + 4x + 3 = 0
x = [ - 4 ± √ (16 - 24 ) ] / 4
x = [ - 4 ± √ (- 8) ] / 4
x = [ - 4 ± i √ (8) ] / 4
x = [ - 4 ± i 2 √2 ] / 4
x = [ - 1 ± i (1/2)√2 ]

Question 2
2x (x - 16) = 0
x = 0 , x = 16

Question 3
2x² - 8x = 3 - 3x
2x² - 5x - 3 = 0
(2x + 1)(x - 3) = 0
x = - 1/2 , x = 3

1. Use the quadratic formula where a=2, b=4 and c=3.

2. Divide throughout by 2 to leave you with:
x^2 - 16x = 0
Then take out a factor x so it looks like this:
x ( x - 16 ) = 0
So either x = 0 or x = 16.

3. Remove the brackets to leave:
2x^2 - 8x = 3 - 3x
Then subtract (3 - 3x) from either side to leave:
2x^2 - 5x - 3 = 0
Now factorise so it is in the form:
(2x + 1)(x - 3) = 0
So either x = -1/2 or x = 3.

Hope that helps!!

2x^2+4x+3=0
x=(-4+-sqrt(16-4*2*3))/2*2
x=(-4+-sqrt(-8))/4
x=-1+-isqrt(2*4)/4
x=-1+-1/2 * sqrt(2)

2x^2-32x=0
2x(x-16)=0
x=0
x=16

2x(x-4)=3(1-x)
2x^2-8x=3-3x
2x^2-5x-3=0
x=[5+-sqrt(25+4*2*3)]/(2*2)
x=[5+-sqrt(49)]/4
x=(5+-7)/4
x=12/4=3
x=-2/4=-1/2

I hope I was helpful. Good luck!

start like this
2x^2 +4x+ 3=0

useing the quadratic equation

[-b +/- sqrt(b^2+4ac)]/a

answer
-2 +/- sqrt (10)