Quadratic Equations 2x^2-x-6=0?

Question

Complete the square.

Everytime I do problems like this or anything that results in a factor with a fraction I get stumped for like 5-10 mins. When it comes to test taking time that'll consume the whole hour of the test, especially with 40 questions!

I understand te concept but I have two questions:
a. Whats the answer
b. How do I get better at factoring out large numbers and fractions

Answer 1

One way to do this problem that does not involve completing the square is to do the following:
Take the coefficient of the x^2 terms and multiply by the constant term.
So 2 times -6 = -12.
Now try to find two numbers that multiply to -12 and add to -1 (-1 being the coefficient of the middle x term)

-4 times +3 = -12
And
-4 plus +3 = -1

So you can rewrite the original polynomial to
2x^2-4x+3x-6 = 0
Factor by grouping the first two terms together and then the second two.
2x(x-2) +3(x-2) = 0
The stuff in parenthesis is the same so you can factor that out and get
(x-2)(2x+3) = 0
So now set each of the factors equal to zero and solve for x.

The best way to get better at these problems is to just keep practicing with them.

Hope this makes sense. Good Luck.

Answer 2

Ax^2 + bx + c= 0

2x^2-x-6=0
x^2 - 1/2 x -3=0
x^2 -1/2 x = 3
(-1/2 *1/2)^2 =1/16 (this part is called completeing the square, you take "b" and multiply it by 1/2, and then square the answer)
x^2 - 1/2 x + 1/16 = 3 +1/16
x^2 -1/2 x + 1/16 =49/16
(x-1/4)^2 = 49/16
(x-1/4) = +- (7/4)
x= 1/4 +- (7/4)
x = 2, -6/4


if you dont understand what i said, or you need further help, e-mail me @ cancer_undercover (of course at yahoo)
good luck

Answer 3

If all else fails, memorize the quadratic equation.
[-b±sqrt(b^2-4ac)]/2a

If b^2-4ac>0, there are no real zeros, so don't kill yourself if it is unfactorable. Here, (-1)^2 - 4(2)(-6)=49 with its square root, 7.
[-(1)±7]/2(2)
x= -8/4 = -2; x = 6/4 = 3/2

In this case,
2x^2 - x - 6 = 0
x^2 - 1/2x -3 = 0
Inspection suggests ±3(1/2) as a factor. You could try both and see if one works, but it involves a lot of trial and error. If your told to factor by inspection, simplify as much as you can and relate b and c. In real life, it's usually to painful.

Answer 4

I think you should ask "How do I complete the square to find the solution to this equation"?

2x^2 - x - 6 = 0
2x^2 - x = 6
x^2 - (1/2)x = 3
x^2 - (1/2)x + 1/16 = 3 + 1/16
(x - 1/4)^2 = 49/16

x -1/4 = +-7/4
x = 1/4 +- 7/4

x = 2 and - 3/2

Answer 5

THIS IS A SIMPLE WHICH CAN BE MASTERED BY PRACTISE AND DONT LOSE HOPE INCASE YOU AR GETTING AN ANSWER IN FRACTIONS.WHO KNOWS ! YOU MIGHT BE CORRECT
2x^2-x-6=0 by splitting the middle term we get
2x^2-4x+3x-6=0
taking first two and the second two we get the following step
2x(x-2)+3(x-2)=0
i.e. (2x+3)(x-2)=0
case1
x-2=0
x= 2

case2
2x+3=0
2x= -3
x= -3/2

Answer 6

2x^2 - x -6 =0
2x^2 -4x +3x - 6 =0
2x ( x - 2) + 3 ( x - 2 ) = 0
(2x + 3 ) ( x - 2 ) = 0
2x +3 = 0, 2x = -3, x = -3 /2
x - 2 =0 , x =2.

What I would suggest is that you practice with a lot of numbers.
Once you get the hang of it , what I did was write the prime factors of the product all together and try to arrange them to get the sum required.

Answer 7

2x^2-x-6=0

2x - 3 3x
x - 2 4x

(2x-3)(x-2)=0

2x-3=0
2x=3
x=3/2

x-2=0
x=2

Answer 8

2x^2-x-6=0
2(x^2-1/2 x -3)=0
2( x^2 -1/2 x + 1/4 -1/4 -3)=0
2(x-1/2)^2 -1/2 -6=0
2(x-1/2)^2 -13/2=0