2y^2-3y-6=0?

y = [ 3 ± √(9 + 48) ] / 8
y = [ 3 ± √(57) ] / 8

2007-09-17 19:51:46 · answer #1 · answered by Como 7 · 1⤊ 0⤋

First of all don't be thrown by the fact that the equation has y in it! Replace y with x and you will probably immediately recognise this as a quadratic equation. So, you need one of the methods for solving quadratic equations.
This particular equation does not have any whole number solutions so you best use the quadratic formula:

x =[ -b + or - sq.rt (b^2 - 4ac)]/2a

Compare your equation with a general quadratic expression:

ax^2 +bx + c and you have a=2, b=-3 and c =-6, and of course y =x. So using the formula you get:

y =[ -(-3) + or - sq. rt (-3^2 - 4x2x-6)]/2x2

y = [3 + or - sq.rt 57]/4

If the question asks you to give exact solutions then these are the two answers. If not you can use a calculator to get
y = 2.637 and y = -1.137 (to 3 dp)

2007-09-17 00:55:02 · answer #2 · answered by Anonymous · 0⤊ 1⤋

before you start solving any quad prob, u should consider these 3 cases.
case 1 :
b^2-4ac>0 (u can factorize it but will hv DIFF root)
case 2 :
b^2-4ac<0 (u CAN't factorize it)
case 3 :
b^2-4ac=0 (u can factorize but will hv same root)

what is a, b, c?

ax^2+bx+c=0

so, in ur problem, identify a,b,c.
i believe that it can't be factorize. so, what to do?

well, use the formula of :

x = [-b±sqrt (b^2-4ac) ] / 2a

after u subtitute all the value :
~in your case~

y = [-(-3)±sqrt(9-4(2)(-6)] / 2(2)

u must solve the square root first before u move on!!!!

y1 = 3 - (sqrt 57) / 4
y2 = 3 + (sqrt 57) / 4

that's the answer~~
hope u'll get it~~

2007-09-16 14:54:35 · answer #3 · answered by suzie s 1 · 0⤊ 0⤋

did you mean this 2y^2-3-6=0?

2007-09-16 14:43:09 · answer #4 · answered by william l 4 · 0⤊ 0⤋

in the beginning ,for fixing a equation with 2 variables,you decide on 2 equations. whether you have a single equation you are able to remedy via trial and blunder technique that is to make certain which fee of x and y siuts the equation. right here that is x=-6;y=2

2016-12-26 14:30:20 · answer #5 · answered by ? 3 · 0⤊ 0⤋

umm ok, Im not sure what you need:

so 2y² - 3y - 6 = 0

then 2y² - 3y = 6

factored into y(2y-3) = 6

then into: 2y-3 = 6/y

thats about it...

2007-09-16 14:41:38 · answer #6 · answered by Anonymous · 0⤊ 0⤋

It cant be factored, so use quadratic formula,
y = [-b +/- sqrt(b^2 - 4ac]/2a

y = [3 +/- sqrt(9 + 48) ]/4
y = [3 +/- sqrt(57) ]/4

2007-09-16 14:38:00 · answer #7 · answered by Anonymous · 2⤊ 0⤋

y=(sqrt(57)+3)/4=2.6374586088177

or

y=(3-sqrt(57))/4=-1.1374586088177

prob set for y

x=0

2007-09-16 17:58:39 · answer #8 · answered by Anonymous · 0⤊ 0⤋

You can use the quadratic formula to solve this:

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

using this gives y = 2.637 or y = -1.137

2007-09-16 20:47:49 · answer #9 · answered by tinned_tuna 3 · 0⤊ 0⤋

Give it a rest mate.

2007-09-16 14:37:00 · answer #10 · answered by Anonymous · 0⤊ 1⤋

11 answers