Can someone please show me how to solve 2x 2 - 6x-10=0?

Question

To any TEACHERS, FELLOW STUDENTS OR MATH HOUNDS, PLEASE HELP. Its finals time and I would really appreciate any help, even referrals to any good math help sites. I need help Completing the Square.

Answer 1

2x^2-6x-10=0
dividing by 2
x^2-3x-5=0
usingthe quadratic formula
x=[3+/-rt(9+29)]/2
=[-3+/-rt29]/1

Answer 2

To understand completing the square, consider the perfect square trinomial, (a+b)^2 = a^2 + 2ab + b^2. (FOIL it out, don't simply "distribute" the exponent.) You see there is a square on each end, and twice the product of the two roots in the middle. When you complete the square, you want to take an arbitrary quadratic expression and "force" it into the form of a perfect square trinomial, with a square on each end and twice the product of the two roots in the middle. That's why you carry out the following steps:
1. Bring the constant to the other side by adding its opposite. Here, that means
2x^2 - 6x = 10
2. If there is a lead coeffieicnt, factor it out.
Here, factor out the 2 and get 2(x^2 - 3x) = 10
3. Take half the second number, square it, and add it to each side. When you add it to the right side, remember that you actually added the product of the number you added inside, with the lead coeffieint that you factored out. So here you'd do...
2(x^2 - 3x + (3/2)^2) = 10 + 2(3/2)^2 =
2(x^2 - 3x + 9/4) = 10 + 18/4 = 10 + 4 1/2 = 14 1/2
4. divide each side by the lead coeffiecnt that you factored out (some people do this first but I do it here cos it's cleaner)
x^2 - 3x + 9/4 = (14 1/2) / 2 = 29 /4
5. Factor the left side, which is now a perfect square trinomial. To do tihs, write, in parenthesis, the square root of the first and last terms, with the first sign in between. here the first sign in between is - and the square root of 9/4 is 3/2 so you get
(x + 3/2)^2 = 29/4
6. Here's the magic part. Take the square root of each side. On the left, that means simply dropping the exponent. On the right, remember to include the + / - so you get your two answers like a quadratic equation is supposed to have...
(x+3/2) = +/- sqrt (29/4)
Drop the parenthesis on the left, you no longer need them...
x + 3/2 = +/- sqrt (29/4)
Finally bring the constant over to the other side by adding its opposite, this leaves x by itself on the left and the answer on the right...
x = -3/2 +/- sqrt (29/4)
which simplifies and rearranges to (-3 +/- sqrt 29)/2
If this looks like the answer you'd get from the quadratic formula, that's because the quadratic formula is derived by completing the square on the general case ax^2 + bx + c = 0.
So why do all this? Why not just use the quadratic formula?
Well, after you get past this, you usually will. But completing the square is a procedure that will come in handy in other areas of math, especially if you get into trig or calculus, so they want you to learn how to do it at this time.
Good luck on your exam!

Answer 3

Example: Solve for x: (x + 5)2 = 18.

= .
| x + 5| = 3.
x + 5 = 3 or x + 5 = - 3.
x = 3 - 5 or x = - 3 - 5.
Since we cannot take the square root of a negative number, there are often numbers which appear to be solutions but do not actually make the equation true. For example, we get two solutions when we solve = - 2x:

()2 = (- 2x)2.
6x + 10 = 4x2.
4x2 - 6x - 10 = 0.
2(2x2 - 3x - 5) = 0.
2(x + 1)(2x - 5) = 0.
x = - 1 or .

We can plug -1 in for x in the original equation to check that it makes the equation true:

= - 2(- 1)?
= 2? Yes.
True.

However, when we plug in for x:

= - 2()?
= - 5? No.
False.

is an extraneous solution, and the only solution to = - 2x is x = - 1.

Answer 4

Since you specified completing the square:

2x² - 6x - 10 = 0

Now, this problem could be made easier by dividing both sides by 2 first:

x² - 3x - 5 = 0

Then, to complete the square, first get all the x's on the left, and the numbers on the right:

x² - 3x = 5

Then (and this is the tricky part), you take the number in front of the x (3 in this case), divide it by the number in front of the x² (1 in this case), divide it again by 2 (3/2), and square it (3/2 * 3/2 = 9/4).

Add that number to both sides:

x² - 3x + 9/4 = 5 + 9/4 = 20/4 + 9/4 = 29/4

Now the left side of the equation is a perfect square: it's (x - 3/2)².

(x - 3/2)² = 29/4

Then take the square root of both sides, remembering that the square root can be either positive or negative:

x - 3/2 = ±√(29/4) = ±√29/2

And then get x by itself to get your answer:

x = 3/2 ±√29/2 = (3 ±√29)/2

Follow that?

Answer 5

a million) (5x + 3) - (x - 2) + (2x - a million). eliminate the brackets and modify the indications contained in the 2d brackets. 5x + 3 - x + 2 + 2x - a million. integrate like words. 6x + 4. component out 2. 2(3x + 2). answer. 2) 4x^2 - 12x = 40. Subtract 40 from both area of the equation. 4x^2 - 12x - 40 = 0. Divide by by 4. x^2 - 3x - 10 = 0. component into 2 binomials. locate 2 numbers that once elevated at the same time = - 10 and at the same time as extra at the same time = - 3. because you've a - 10, between the numbers must be - and one +. because you've - 3, the finest of both numbers must be -. attempt - 5 and + 2. (x - 5)(x + 2). answer

Answer 6

2x^2 -6x -10 = 0
Divide by 2 getting
x^2 -3x -5 = 0
x^2-3x = 5
Now take the number multiplying x which is 3 divide it by 2 getting 3/2 and then square it getting 9/4. Now add it to both sides of the equation getting:
x^2-3x+9/4 = 5+9/4 =29/4
Now we have completed the square so
(x-3/2)^2 =29/4
Now take square root of both sides getting:
x-3/2 = sqrt(29/4) = +/- 1/2sqrt(29)
x= 3/2 +/- 1/2sqrt(29)

Hope that helps you to understand completing the square.

Answer 7

Hey there, lucky!
I am not quite sure about what your problem is, I am having some trouble reading it. But, I suggest that you try solving using the quadratic equation. It has been a long time for me since I have done that math, sorry! But I hope that this was helpful anyways! Just don't stress out about your test and you will do fine. Make sure that you get enough sleep the night before so that your mind is at peak performance, ok? You owe it to yourself to do a good job! And you know that you CAN do it, so just stop worrying, ok? You'll do fine if you have been putting the effort into it!

Answer 8

2x^2 - 6x - 10 = 0
is easiest solved if you divide through by 2, giving
x^2 - 3x - 5 = 0
Completing the square,
x^2 - 3x + (3/2)^2 - (3/2)^2 - 5 = 0
(x - 3/2)^2 = 5 + 9/4 = 29/4
x - 3/2 = ± (1/2)√29
x = 3/2 + (1/2)√29, 3/2 - (1/2)√29

Answer 9

Can someone please show me how to solve 2x² - 6x - 10 = 0

2x² - 6x - 10 = 0
So x² - 3x - 5 = 0

This will not factorise easily so use the formula:

x = (- (-3) ± √((-3)² - 4*(1)*(-5))/(2*1)

= ½(3 ± √29)

≈ -1.1926, 4.1926

Answer 10

First you need to solve the like variables, in other words
2x-6x+2-10=0
-4x-8=0

Then get x on one side and make it positive

-4x+4x-8=0+4x
-8=4x

divide and make x=1

-8/4=4x/4
-2=x
x=-2

I hope that was the problem, it should be algebra 1