completing the square?

Question

can someone show me step by step how to solve this by completing the square:

2x^2 - 7x - 15 = 0

Answer 1

Divide through by 2:

x^2-(7/2)x-(15/2)=0
x^2-(7/2)x=(15/2)

Add half the middle term squared to both sides
(1/2)(-7/2)=-7/4 squared is 49/16

x^2-(7/2)x+49/16=(15/2)+49/16
(x-7/4)^2=169/16

Take square root of both sides

x-7/4=13/4 or x-7/4=-13/4

Solve for x and you're done.

Answer 2

This is a rough one.
Add 15 to both sides so that everything with an x is together.
2x^2 - 7x = 15
Divide both sides by 2

x^2 -7/2 x = 15/2
Add something to both sides so that the left side is a perfect square. The trick; take half of -7/2 and square it.
-7/4 squared = 49/16 add to both sides.

x^2 -7/2x + 49/16 = 15/2 + 49/16
factor (x-7/4)^2 = (120 + 49)/16 = 169/16
Take the square root of both sides
(x - 7/4) = 13/4 or -13/4
Add 7/4 to both sides 20/4 or -6/4
Plug 5 and -1 1/2 into the original expression to see if they work.

Answer 3

To complete the square follow the steps:

2x²-7x-15 = 0
2x²-7x = 15
Let the coefficient of x² equal to 1:
x²-(7/2)x = 15/2
from here you should find a number that is equal to the half of (7/2) and squared:
That number is (7/2)(1/2) = 7/4
and squared is : (7/4)²

Add the number to both sides of the equation:
x²-(7/2)x + (7/4)² = 15/2 + (7/4)² (you completed the square in the left side in this precise moment!!)

You just have to factorize the square:

[x-(7/4)]² = 15/2+49/16

Finally you get:

[x-(7/4)]² = 169/16

That's it!

Good luck!

Answer 4

2xexp2 -7x -15 =0

Divide by 2 to eliminate coefficient to xexp2 term

xexp2 - 7/2x - 15/2 = 0

Move -15/2 to other side of = by addition property

xexp2 - 7/2x = 15/2

halve and then square coefficient to x term

(-7/2) divided by 2 = (-7/4) ; squared = 49/16

Add this term to both sides of equation

xexp2 -7/2x + 49/16 = 15/2 + 49/16

factor left side of equation (use halved and then squared coefficient to x term): (x-7/4)exp2

Now combine terms on right side of equation: 120/16 + 49/16 = 169/16 = (13/4)exp2

Some teachers want you to move all terms to one side, so can use property of subtraction to move over if you need to solve for zero.

Answer 5

I dont know what completing the square means bucause I do my maths through Irish, but this is what I'd do...

2x^2-7x-15=0
(2x+3)(x-15)=0
2x=-3 x=5
x=-3/2

x={-3/2, 5}

Answer 6

Professor maddie shows you correctly to make the left side of the equation into a square and the right side into a real number.
The Object of the exercise is to make the equation into the form:
xsquared +2Bx + Bsquared = N
(x + B)squared = N
When she added "one half the middle term squared"
2B would be the "middle term".
(One half of 2B) squared would be Bsquared and
Then take the square root which would be positive and negative which give you two equations to solve for x.

Answer 7

I am in Algebra II now and we just did this and I dont know if these answers are actually answering your questions, but I cant give you a step by step thing but here is the theorem:

To complete th square on x^2+bx, add (.5b)^2. x^2+ bx + (.5b)^2 = (x + .5b)^2 = (x + b/2)^2

Also you dont divide by 2 to get rid of it you would do something like 2(x) and so when you add (.5b)^2 you would really add ((.5b)^2)/2


Sorry I couldnt help you more.

Answer 8

2x^2-7x-15=0
(2x+3)(x-5)=0
so now we have 2 separate answers for x:
2x+3=0
2x=-3
x=-3/2

and x-5=0
x=5

by the way, professor maddie gave us all thumbs down, but her math is not completing the square

Answer 9

(x+a)(x+b)=x^2+(a+b)x+ab
you have to find the combination...

2*(-5)+3*1=-7
3*(-5)=-15
(2x+3)(x-5)=0