Word Problems(might involve factoring)

[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.

 

[Help_Me_Please]

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

 

[Help_Me_Please]

n x (n+1) = 56
its 7 and 8 .. but I can't seem to remember how to get the answer with the equation

nx (n+2) = 35

5 and 7 =)

 

[Hardeep]

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

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

 

[Help_Me_Please]

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 =)

 

[Help_Me_Please]

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

 

[Hardeep]

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

I see now, thanks fo rthe help

 

[Hardeep]

ok I have just one more;

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

 

[dextercioby]

ok I have just one more;

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

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.