2x^2-x-28=0?

Question

I need help how to do this quadratic equation. I need step by step information to make sure that I under the problem.

Thanks

Answer 1

(2x + 7) (x - 4) = 0
x = - 7 / 2 , x = 4

Answer 2

It does not appear to be one of those that is nice and easy to factor, so you are left with two options. You can either complete the square or just plug the coefficients into the quadratic formula.

Here are instructions for completing the square:
1-Move the constant term over to the right hand side:
2x^2-x=28
2-Divide out by the coefficient of x^2 so that x^2 will have a coefficient of 1 as a result:
x^2-(x/2)=14
3-To find the correct value that will complete the square, divide the coefficient of x by 2 and then square it. Therefore in this case, it would be 1/2 of 1/2 or 1/4 and then squared would be 1/16. Now you need to add this number to both sides of the equation:
x^2-(x/2)+1/16=14+1/16
4-Factor the right side into a perfect square. It will always be the square root of the first term and the square root of the third term. The sign between the terms should be the same as the sign between the 1st and 2nd term in the original.
(x-1/4)^2=14+1/16
5-Simplify the right side. For this problem, just express the whole number 14 in terms of 16ths and add the 1/16 to it.
(x-1/4)^2=225/16
6-Take the square root of both sides. Don't forget that when you take the square root of both sides of an equation, you need to express the right side as +/-
x-1/4= +/- 15/4
7-Add 1/4 to both sides and you have your solutions.
x= 15/4+1/4 =16/4=4
x= -15/4+1/4= -14/4 = -7/2

x=(4, -7/2)

Another approach is to use the quadratic formula. In essence it is the same thing as a standard form quadratic equation converted into a formula.
A quadratic equation in standard form is as follows:
ax^2+bx+c=0 where a, b, and c are the coefficients.
If you were to complete the square for this general equation, you would end up with the following:

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

Therefore, if you were to use this approach for your example, you would plug in 2 for a, -1 for b and -28 for c and solve it.

Answer 3

2x² - x - 28 can be factored down to (2x + 7)(x - 4). After factoring, you are left with x = 4 or 2x = -7, or x = -7/2.

Or, you could use the quadratic formula to get your answer, which says that:

x = [-b ± √(b² - 4ac)] / 2a.

In this case, a=2, b=-1, and c=-28. Putting those in the formula, you get,

x = [-(-1) ± √((-1)² - (4 * 2 * -28)] / (2 * 2).
x = [1 ± √(1 - (-224)] / 4.
x = (1 ± √225) / 4.
x = (1 ± 15) / 4.
x = 16/4 = 4, or x = -14/4 = -7/2.

Answer 4

Hi,

I' m going to explain factoring trinomials the way that I teach it, which is probably different from what you've done. Try to follow the steps. This method will work on any trinomial.

2x² - x - 28 Look for a GCF None this time. If there was one, factor it out. Then temporarily start both parentheses with the first number and variable.
(2x.......)(2x..........) First sign goes in first parentheses.
(2x..-....)(2x..........) Product of signs goes in 2nd parentheses.
(2x..-....)(2x...+.....) <== plus is because neg x neg = positive

Now multiply your first and third numbers together. Ignore their signs - you've already done them. 2 x 28 = 56 So, out to the side list pairs of factors of 56.

56
------
1, 560
2, 28
4, 14
7,8

Now you want to pick which factors go in your parentheses, using these rules:

If the signs you put on your parentheses are the SAME, find the factors that ADD up to the middle number in the problem.
If the signs you put on your parentheses are the DIFFERENT, find the factors that SUBTRACT to the middle number in the problem. (Note you look at signs in the parentheses, not in the problem.)

(2x.-....)(2x.+.....) Your signs are different, so you want to subtract factors to get 1. Those factors are 7 and 8. When you put the numbers in the parentheses, the bigger number is pushy - it always goes first. So your factors now are:
(2x.-.8)(2x.+.7)

Now you have to reduce either or both parentheses by dividing each parentheses' terms by the largest possible divisor. In our problem, the first parentheses is divisible by 2, but the second parentheses does not reduce.

(2x.-.8)(2x.+.7)
---------
....2..... This reduces to your final factors of

(x.-.4)(2x.+.7)

Since this was set equal to zero in the original problem, you now have

(x.-.4)(2x.+.7) = 0 as your equation. Set each factor equal to zero at solve them for your answers.

x - 4 = 0
x = 4 <== 1st answer

2x + 7 = 0
2x = -7
x = -7/2 <== 2nd answer

Your answers are 4 and -7/2.

The factoring explanation will work for any quadratic equation that is factorable.

I hope that helps!! :-)

Answer 5

You've got a negative sign on the end, so the two operators in the factors are going to be a plus and a minus. That means the factors are in the form of

(2x + ?)(x - ?) or (2x - ?)(x + ?)

We need to find two factors of 28 that also give us the middle term of -x. The combination that works is:
(2x + 7)(x - 4)

Answer 6

2x^2 - x - 28 = 0
(x - 4) (2x + 7) = 0

x - 4 = 0, x = 4
2x + 7 = 0, x = - 7/2

Answer: x = 4 and x = - 7/2

Answer 7

2x^2-x-28 = 0
(2x+7) (x-4) = 0 ( I factored it using foil method )

what values of x equals zero x= -3.5 and x = 4

Answer 8

2x^2-x-28=0
or 2x^2+7x-8x-28=0
or x(2x+7) -4(2x+7)=0
or (x-4) (2x+7) = 0
So either x-4=0 or 2x+7=0
So, x= 4 or -7/2

Answer 9

2x^2-x-28=0

Factor:
(x-4)(2x+7)=0
Check: (x)(2x) = 2x^2, -4(2x) = -8x, 7(x) = 7x and -8x +7x = -x, -4(7) = -28

x-4=0, 2x+7=0 (this is also 2x=-7, divide both sides by 2)

x=4, x=-7/2

Answer 10

2x^2-x-28=0
x^2-28=0
x^2=-28
x= square root of -28