# Word Problems(might involve factoring)

1. Jan 2, 2005

### Hardeep

Im having a bit of trouble with these:

1). Seven less than 4 times the square of a number is 18. Find the number.

2). Find two consecutive integers whose product is 56.

3). Find two consecutive positive odd integers whose product is 35.

2. Jan 2, 2005

4n^2 - 7 = 18
4n^2 = 25
n^2=6.25
n=2.5

3. Jan 2, 2005

n x (n+1) = 56
its 7 and 8 .. but I cant seem to remember how to get the answer with the equation

nx (n+2) = 35

5 and 7 =)

4. Jan 2, 2005

### Hardeep

I know that why, but we have to factor to get it. Like

x^2 - x = 6
x^2 - x - 6 = 0
(x + 2)(x -3)

x= -2 OR x= 3

5. Jan 2, 2005

4n^2 - 25 = 0
4n^2 -10n + 10 n -25 = 0
2n (4n^2 - 10n) + ( 10n - 25) 5
2n - 5 + 2n- 5 2n+5

2n-5 and 2n+5 =)

6. Jan 2, 2005

what I did was I took the 25 and multiplied it by the 4 and got 100 ... then I thought of factors of 100 that would add to give me 0 .. and just put it in like it was part of the problem

7. Jan 2, 2005

### Hardeep

I see now, thanx fo rthe help

Last edited: Jan 2, 2005
8. Jan 2, 2005

### Hardeep

ok I have just one more;

4). The sum of the sqaures of two consective integers is 41. FInd the integers.

9. Jan 2, 2005

### dextercioby

Call them 'a' and 'b'.Then
$$a^{2}+b^{2}=41$$.Since 'a','b' are integers,then both 'a' and 'b' in modulus must be smaller or at most equal to 6,as 6 is the greatest integer whose square is less/equal to 41.
The only possible sollutions to the eq. are (4,5),(-4,-5) and viceversa.The mixed combinations don't have consecutive integers.

Daniel.