What are the answers to these algebra problems?

Question

thanks for the help!

1) find two integers whose differences is 23 and whose product is negative 120.

2) the sum of the square of two consecutive even integers is 100. find the integers.

3) when one integer is added to the square of the next consecutive integer, the sum is 55. Find the integers.

thanks again for the help

Answer 1

1) 15 and -8

2) 36+64 = 100, s0 6 and 8 are two such numbers

3) 6+7^2 = 6+49 = 55, so 6 and 7 are two such integers

Answer 2

Finding integer solutions to equations is called Diophantine analysis. It helps to practice to make discrete math easier later (and some aspects of statistics).

1) Let a be the greater of the two numbers:
a - b = 23
ab = -120

From second equation:
Only one of a or b is negative.

From first equation:
b cannot be less than -22
a cannot be more than +22

At this stage, because there are only a small number of possible combinations of a and b, you could simply try them out

a , b , ab , ?
1, -22, -22, no
2, -21, -42, no
3, -20, -60, no
etc.

We could use a few shortcuts, such as
ab = -120 tells us that one of a,b has to be divible by 5

0, -23, 0, no
3, -20, -60, no
5, -18, -90, no
8, -15, -120, yes
10, -13, -130, no
13, -10, -130, no
15, -8, -120, yes
18, -5, -90, no
etc.


2) b is smaller than a
a = b+2
a^2 + b^2 = 100

From the second equation, we know that a positive b has to be less than the square root of 50 (since a is bigger than b, then b^2 has to be less than half of 100). So, we only need to check values less than 7.
However, because the square of a negative number is positive, we also have to check negative values down to -8 (for b to be -7).

We also know that a has to be greater than SQRT(50). Otherwise, if both b^2 less than 50 AND a^2 less than 50, then they cannot add up to 100.

That leaves only one positive choice:
b = 6 and a = 8

and we check the negative choice:
b = -8 and a = -6
b^2=64 a^2=36 sum=100

3) Again, b is smaller than a
a=b+1
b + (b+1)^2 = 55

b cannot be 0 or -1.
(Many problems are either solved or made easier by checking 0 and numbers very close to 0 right off the start).

If b is positive, then (b+1)^2 has to be less than 55

The first square below 55 is 49 (= 7 squared)

b, b+1, sum, ?

6 , 7 , 6 + 49 = 55, yes
5 , 6 , 5 + 36 = 41, no
4 , 5 , 4 + 25 = 29, no
(safe to stop)

If b is negative, then we have to be a little more careful, as the negative b may offsett a slightly too large (b+1)^2

-10, -9, -10 + 81 = 71, no
- 9 , -8, -9 + 64 = 55, yes
-8 , -7, -8 + 49 = 41, no
-7 , -6 , -7 + 36 = 29, no
etc.

Answer 3

1)
a - b = 23
a.b = -120

a = 23 + b

(23 + b)b + 120 = 0
23b + b² + 120 = 0
d = 23² - 4.1.120
d = 529 - 480
d = 49

b = (-23 +/- \/49) : 2
b' = (-23 + 7) : 2
b' = -16 : 2 = -8
b" = -30 : 2 = -15

Answer 4

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