Quadratic - Completing the square problem.?

Question

y^2 - 7y = -10.

I am supposed to divide -7 by 2 (or in half).
and then square that.
well -7 divided by 2 is -3.5
-3.5 * -3.5 = 12.25
then I am supposed to add 12.25 to both sides. well should I round it off to 12?
what did I do wrong. I always have a hard time figuring these when the coeffient of the variable is an odd number.
thanks if you can help me.

Answer 1

Do not round off.
You should have y^2-7y +12.25 = -10 +12.25
(y-3.5)^2 = 2.25
y-3.5 = +/- 1.5
y = 3.5 +/- 1.5
y = 5 and 2

Answer 2

You have done nothing wrong. Let me carry on from where you left off.
y^2-7y+12.25=-10+12.25
(y-3.5)^2=2.25
Taking the square root of both sides gives
(y-3.5)=+ or - 1.5
y=3.5+1.5 or 3.5-1.5
y=5 or 2
Which are the correct answers.

As you progress, you will ,I think ,grow more comfortable with fractions . One thing for sure-they avoid decimals! Lastly, you are not allowed to "round off" unless you are estimating, or giving approximations. Rounding off because you don't like decimals is verboten!!
I believe you will shortly be learning the Quadratic Formula. It begins with the General equation for any quadratic, namely ax^2 +bx+c=0, and says to find the x values for ANY quadratic, the formula
x={-b+or-rt(b^2-4ac)}/2a works every time.

Getting to that formula involves 3 steps:
Step1-make the coefficient of the x^2 term +1
Step2-move the constant term to the right side
Step3) -complete the square.

I hope your mastery of completing the square lets you follow the formula's development when the time comes.

Answer 3

what u did wrong was rounding it off, you are NOT to do that. Remember that in math you just can't go around changing numbers just because you feel like it.

y^2-7y=-10
7/2=3.5
3.5^2=12.25

then
y^2-7y+12.25=-10+12.25
y^2-7y+12.25=2.25

then
(y-3.5)^2=2.25

thats the answer. Oh and y=2!

Answer 4

You don't want to round these; the answer won't come out right. Instead, you might want to try leaving decimals in fraction form. It makes things easier.
To solve:
y^2-7y+(7/2)^2 = -10+(7/2)^2
y^2-7y+(49/4) = -10+(49/4)
(y-(7/2))^2 = (-40/4) + (49/4)
(y-(7/2))^2 = (9/4)
square root both sides:
(y-(7/2)) = +/- 3/2
add (7/2) to both + 3/2 and - 3/2:
y = 10/2 and y = 4/2
y = 5 and y = 2

Answer 5

hmm..
adding 12.25 to both sides will get
y^2 - 7y + 12.25 = -10 + 12.25

so;
(y - 3.5)^2 = 2.25
simplifying, by getting the square of both sides,
y - 3.5 = 1.5
transposing:
y = 3.5 +/- 1.5

so the answers could be:
y = 2, y = 5

Answer 6

You are not applying the quadratic equation

First you must rearrange your equation so it looks like this

ay^2 + by + c = 0

you must get zero on the other side

so your equation becomes

y^2 -7y +10 = 0

Can also be written as

(1)y^2 + (-7)y + (10) = 0

So as you can see

a= 1
b=(-7)
c= 10

Now we must plug these into our quadratic equation.

(roots) = [(-b)(+/-)SQRT((b^2) + 4ac)]/(2a)

(roots) = [(-(-7))(+/-)SQRT((-7)^2 + 4 * 1 * 10)]/(2 * 1)

(roots) = [(7) (+/-) SQRT((49) + 40)]/2

(roots) = [7 +/- SQRT(9)]/2

(roots) = [7 +/- 3]/2

(roots) = 4/2 and 10/2

So there are two roots to the equation, 2 and 5

plug back into the original equation as a check

(5)^2 - 7 * (5) = -10

and

(2)^2 - 7 * (2) = -10

So they check out right

Good Luck

Answer 7

y^2 - 7y + 12.25 = -10 + 12.25
(y - 3.5)^2 - 2.25 = 0