[Help] Algebra Math Problems HELP NEEDED NOW!

Algebra Math Problems HELP NEEDED NOW!

Please help answer these 2 Algebra 1 questions!

1) Eight less than nine times a number is the same as the square of the number. Find the number.

2) Find the consecutive odd integers such that the square of the third, decreased by the first, is 46.

PLEASE SHOW WORK!!! THANKSS!!

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Originally Posted by MagixAries

Please help answer these 2 Algebra 1 questions!

[B]1) Eight less than nine times a number is the same as the square of the number. Find the number.

PLEASE SHOW WORK!!! THANKSS!!

I only got the 1st one. And even that I'm not sure about.

(I just guessed and checked)

(G x 9)-8 = G
(1 x 9)-8 = G
9-8= 1

Square root of 1 is 1.

wolframalpha.com

SOLVES ALL MATH PROBLEMS

Thanks everyone! But can anyone do it algebraically? I really need some help on this! xD

Can someone please help? I need this by tonight midnight!! DX
@Alex @Alen
@Ravallo

Originally Posted by MagixAries

Can someone please help? I need this by tonight midnight!! DX
@Alex @Alen
@Ravallo

I will do this. Don't worry.

---------- Post added at 06:23 PM ---------- Previous post was at 06:05 PM ----------

So the first equation will be
(9x)-8=x^2
We get this since Nine times a # is 9x, eight less than that is 9x-8.
To solve this, you have to first get common terms. So...
We can factor it like a polynomial
-X^2+9X-8=0
From there we can factor the equation...
But the solutions are extraneous, It cannot be factored. (Since you can get -8 from two factors of 9).
The best possible method now is to graph the equation above and find the x-intercepts
when you graph X^2-9X+8 (I multiplied the whole equation by -1 so X^2 wouldn't be negative).
You get two intersections.. one at -10, and one at 1.
CHECK to make sure the solutions are right.
100-90+8 DOES NOT WORK
but
1-9+8 DOES work.

Is this what your teacher wanted you to do?

---------- Post added at 06:28 PM ---------- Previous post was at 06:23 PM ----------

consecutive odd integers are
7,9,11 (for example).
So what we want to know is which consecutive odd integers ( 3 of them) equal 46.
We know we only need 3 integers since in the question it says. Square of THIRD, decreased by FIRST.
Basically all we have to do is find a # (The one we are squaring... the third #)LESS THAN 9, and the same # must be greater than 5. Since 5^2 is 25, which isn't greater than 46..
So the only ODD # we are stuck with is 7 (greatest number).
So the consecutive odds are 3,5,7
7^2 =49
49-3=46
THERE YOU GO

Oh darn, looks like he beat me to the punch. Oh well, hope it helps.

Originally Posted by FeelTheRush

I will do this. Don't worry.

---------- Post added at 06:23 PM ---------- Previous post was at 06:05 PM ----------

So the first equation will be
(9x)-8=x^2
We get this since Nine times a # is 9x, eight less than that is 9x-8.
To solve this, you have to first get common terms. So...
We can factor it like a polynomial
-X^2+9X-8=0
From there we can factor the equation...
But the solutions are extraneous, It cannot be factored. (Since you can get -8 from two factors of 9).
The best possible method now is to graph the equation above and find the x-intercepts
when you graph X^2-9X+8 (I multiplied the whole equation by -1 so X^2 wouldn't be negative).
You get two intersections.. one at -10, and one at 1.
CHECK to make sure the solutions are right.
100-90+8 DOES NOT WORK
but
1-9+8 DOES work.

Is this what your teacher wanted you to do?

---------- Post added at 06:28 PM ---------- Previous post was at 06:23 PM ----------

consecutive odd integers are
7,9,11 (for example).
So what we want to know is which consecutive odd integers ( 3 of them) equal 46.
We know we only need 3 integers since in the question it says. Square of THIRD, decreased by FIRST.
Basically all we have to do is find a # (The one we are squaring... the third #)LESS THAN 9, and the same # must be greater than 5. Since 5^2 is 25, which isn't greater than 46..
So the only ODD # we are stuck with is 7 (greatest number).
So the consecutive odds are 3,5,7
7^2 =49
49-3=46
THERE YOU GO

Thanks, bro, but the 2nd problem I'm not sure....

I start out like this:
x
x+2
x+4

with those 3 terms representing the three consecutive odd integers..but after that I just don't know how to set up the problem at all...any idea?

Originally Posted by MagixAries

Thanks, bro, but the 2nd problem I'm not sure....

I start out like this:
x
x+2
x+4

with those 3 terms representing the three consecutive odd integers..but after that I just don't know how to set up the problem at all...any idea?

Look at Izzy's post. I think he did what you started out to do.

Originally Posted by MagixAries

Can someone please help? I need this by tonight midnight!! DX
@Alex @Alen
@Ravallo

Sorry to tell, but I'm the wrong Alex you are seeking for.

HelpAlgebra Math Problems HELP NEEDED NOW!

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