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3x^2 - 2x = 10

I did.

x = 2 (2)^2 - 4 (3) (10) divided by 2 ( 3)

x = 2 4 - 120 divided by 6

I got 2 + 4.3 divded by 10 equals 0.83

2 - 6 divided by 6 equals 0.6

The final answers I got are 0.83 and 0.6 is that correct, if not could someone please help me.

2007-11-28 15:02:19 · 10 answers · asked by gizzie76 1 in Science & Mathematics Mathematics

10 answers

check your answers. Put those numbers where the X and see if both side equal the same. If they don't then you have the wrong answers.

2007-11-28 15:08:04 · answer #1 · answered by Shanna Saunders 1 · 0 0

The only mistake you made was not to first put it into standard form. You have to have 0 on the right-hand side. It becomes:

3x^x - 2x - 10 = 0

The discriminant becomes b^2 - 4ac = 4 + 4.3.10 = 124

( 2 +/- root(124) ) / 6

= 2.189 or -1.523

Check by substitution - these are correct.

2007-11-28 23:11:26 · answer #2 · answered by Anonymous · 0 0

a x^2 + b x + c = 0

x = [ -b +/- SQRT( b^2 - 4ac ) ] / 2a

In your problem
a = 3
b = -2
c = -10

b^2 - 4ac = 4 - 4(3)(-10) = 4 +120 = 124

SQRT(124) = 2*SQRT(31) = 11.136 (approximately)

x = [2 +/- 2√31 ] / 6

I don't get 0.6 nor 0.83

2007-11-28 23:15:48 · answer #3 · answered by Raymond 7 · 0 0

so the quadratic formula reads x= -b +/- squareroot of b^2 - 4ac all over 2a
so the answers to your problem are [1/3(1+/- squareroot of 31)]

so the answers written out in words because it may be a little unclear, are, one third times one plus or minus the square root of 31.

if you have further questions you can email me, if you have a graphing calculator, its very easy to come up these answers, you just graph the function, and the line y = 0, and find the two intersection points. glad to help

2007-11-28 23:18:28 · answer #4 · answered by P 3 · 0 0

no bc first you have to set the equation to 0
so you get 3x^2-2x-10=0
then plug them in
2+-2^2-4 3 -10 /2 3
2 +- 4+120 /6
2 +- Sr 124 /6

(2+11.14)/6 (2-11.14)/6
2.19 -1.523

2007-11-28 23:09:32 · answer #5 · answered by Briana L 4 · 1 0

you have to set the equation to the quadratic form (ax^2+bx+c=0) first, that is 3x^2 - 2x - 10 = 0. to find for x or the root of the equation, just use the quadratic formula n this website:

http://en.wikipedia.org/wiki/Quadratic_formula#Quadratic_formula

a=3, b=negative 2 and c= negative 10

your answers would be negative 1.52258812 or simply negative 1.52 and 2.1892547 or 2.19 only.

=]

2007-11-28 23:28:46 · answer #6 · answered by dr.psycho 2 · 0 0

3x^2 - 2x = 10 --> 3x^2 - 2x - 10 = 0

D = sqrt(b^2 - 4ac) = sqrt(4 - 4*3*(-10)) = sqrt(4 + 120)
D = sqrt(124) = 2 sqrt(31)

x1 = (-b + D)/2a = (2 + 2 sqrt(31)) / 6 = 2(1 + sqrt(31)) / 6
--> = (1 + sqrt(31)) / 3 = 2.18925478761

x2 = (-b - D)/2a = (2 - 2 sqrt(31)) / 6 = 2(1 - sqrt(31)) / 6
--> = (1 - sqrt(31)) / 3 = -1.52258812

2007-11-28 23:19:49 · answer #7 · answered by Seto 2 · 0 0

did you reright the forumula in ax^2+bx+c?

3x^2 - 2x - 10 = 0

a = 3
b= -2
c = -10

so it wouldnt be - 4 (3) (10), it would be - 4 (3) (-10), which makes it +120

=/

2007-11-28 23:11:10 · answer #8 · answered by Gucci G 2 · 0 0

x= -b +/- sqrt (b^2-4ac)/2a

a=3, b=-2,c=0

x=2+/- sqrt (4)/2(3)

x=2/3, 0

2007-11-28 23:08:47 · answer #9 · answered by Anonymous · 0 2

seems ok

2007-11-28 23:06:07 · answer #10 · answered by Juan O 6 · 0 1

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