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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Thanks everyone! But can anyone do it algebraically? I really need some help on this! xD
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---------- 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
Ixxz (02-06-2013)
Oh darn, looks like he beat me to the punch. Oh well, hope it helps.
Last edited by Ixxz; 02-06-2013 at 06:41 PM.
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