Can anyone help me on these algebra 1 problems.?

Question

1) The product of two consecutive odd whole numbers is 143. find the numbers.


2) Twenty more than the square of a number is twelve times the number. Find the number.\


3)the product of three more than a number times two less than the number is fourteen. Find the number.


4) The sum of the squares of two consecutive odd ositive integers is 202. Find the numbers.


5)The product of two consecutive even whole numbers is 288. Find the numbers.

Please helpme on these problems.

Answer 1

1) x*(x+2) = 143
x^2 + 2x -143 = 0
(x-11)(x+13) = 0
x = 11, -13
y = 13, -11(ans)

2) x^2 + 20 = 12x
x^2 -12x + 20 = 0
(x - 10)(x - 2) = 0
x = 2, 10 (ans)

3) (x + 3)(x - 2) = 14
x^2 + x - 6 = 14
x^2 + x - 20 = 0
(x - 4)(x + 5) = 0
x = 4, -5 (ans)

4) x^2 + (x + 2)^2 = 202
x^2 + x^2 + 4x + 4 = 202
x^2 + 2x - 99 = 0
(x - 9)(x+11) = 0
x = 9, -11 (omit -11 cos question asked for +ve number. so x = 9 and the next consecutive odd number is 11) (ans)

5) x*(x + 2) = 288
x^2 + 2x - 288 = 0
(x - 16)(x + 18) = 0
x = 16, -18
y = 18, -16

Answer 2

I'll explain how to approach the first problem so you can complete the remaining four.

The first problem asks you to find the two consecutive odd numbers whose product is 143.

So what we can do is try to make an equation out of this problem.

Lets call the first odd number, x.
We can call the second odd number, y.

Now we know that the odd numbers are consecutive, which means that y is actually x+2 (an example of how this works is if x = 1, y = 3).

So, now we take what we know, x * y = 143
We can use y = x +2 to substitute for the Y in the above equation.

x*(x+2) = 143

x^2+2x-143=0

Now all you have to do is solve for x,

Answer 3

1) 11 & 13
2) 2 & 10
3) 4 & -5
4) 9 & 11
5) 16 & 18

Use quadratic eqns to solve all these.

Answer 4

1) 11 and 13

2) 2 and 24

3) Don't understand your question

4) 9 and 11

6) 16 and 18

Cmon, you should be able to do these by trial n error !! ;o)

Answer 5

sparkling up for x for the 1st equation: 2x=4+2y, now devide by skill of two the two sides x=2+y, now replace this into equation #2 for x: 3(2+y)+2y=11 and sparkling up for Y, so 6+3y+2y=11 5y=11-6 ----> Y=a million, now replace this for equation #a million and sparkling up for X 2x-2(a million)=4 2x=4+2 X=3 So we arive at X=3, and Y=a million you're able to try this for the two equation, it is not suitable with wich one you initiate up first.

Answer 6

Sorry. I got a D in algebra

Answer 7

Stop trying to get people to do your homework!