Help With Word Problems?

[Hanajima]

#1
Set up an equation and solve the problem.
Find two consecutive whole numbers such that the sum of their squares is 761.

For this problem I set up the equation:

x^2 + (x+1)^2 =761

And I got the answers -20 and 19, however the program I do my homework on is telling me that these are the wrong answers. Where did I go wrong?

#2
Set up an equation and solve the problem.
Suppose that the sum of two whole numbers is 6, and the sum of their reciprocals is 3/4. Find the numbers.

For this problem I did the following:

I found that the two expressions I could use were "x" and "6-x"
I then found the reciprocal of x and 6-x and got

1/x + 1/(6-x)

Where would I go from here though?

 

[Mrspi]

#1
Set up an equation and solve the problem.
Find two consecutive whole numbers such that the sum of their squares is 761.

For this problem I set up the equation:

x^2 + (x+1)^2 =761

And I got the answers -20 and 19, however the program I do my homework on is telling me that these are the wrong answers. Where did I go wrong?

-20 and 19 are NOT two consecutive whole numbers. In fact, -20 is not a whole number at all. It is true that there are two values of x which satisfy your equation, and those values are -20 and 19.

Which of those is a whole number? And if that is the first of the consecutive whole numbers, what would the next be?

Try those answers and see if your homework program likes them better. (A REAL teacher is better than any computer homework program, in my opinion.)

#2
Set up an equation and solve the problem.
Suppose that the sum of two whole numbers is 6, and the sum of their reciprocals is 3/4. Find the numbers.

For this problem I did the following:

I found that the two expressions I could use were "x" and "6-x"
I then found the reciprocal of x and 6-x and got

1/x + 1/(6-x)

Where would I go from here though?

Ok...did you notice this in your problem? It is the basis for the equation you'll need to use:

The sum of their reciprocals is 3/4

So.....
(1/x) + (1 /[6 - x]) = 3/4

Solve the equation.....

 

[Deleted member 4993]

#1
Set up an equation and solve the problem.
Find two consecutive whole numbers such that the sum of their squares is 761.

For this problem I set up the equation:

x^2 + (x+1)^2 =761

And I got the answers -20 and 19, however the program I do my homework on is telling me that these are the wrong answers. Where did I go wrong?

The number you got (-20 and 19) - do you think those are consecutive?

Your set-up is correct - must be making mistakes in algebra or arithmatic.

#2
Set up an equation and solve the problem.
Suppose that the sum of two whole numbers is 6, and the sum of their reciprocals is 3/4. Find the numbers.

For this problem I did the following:

I found that the two expressions I could use were "x" and "6-x"
I then found the reciprocal of x and 6-x and got

1/x + 1/(6-x) = 3/4

What is the LCD - multiply both sides by the LCD and simplify and solve.


Where would I go from here though?

 

[TchrWill]

Set up an equation and solve the problem.
Find two consecutive whole numbers such that the sum of their squares is 761.

For this problem I set up the equation:

x^2 + (x+1)^2 =761

And I got the answers -20 and 19, however the program I do my homework on is telling me that these are the wrong answers. Where did I go wrong?

#2
Set up an equation and solve the problem.
Suppose that the sum of two whole numbers is 6, and the sum of their reciprocals is 3/4. Find the numbers.

For this problem I did the following:

I found that the two expressions I could use were "x" and "6-x"
I then found the reciprocal of x and 6-x and got

1/x + 1/(6-x)

Where would I go from here though?

Let x and (x + 1) be the two numbers.
Then, x^2 + (x + 1)^2 = 761 which reduces to x^2 + x - 380 = 0.
By means of the quadratic equation, x = [-1+/-sqrt(1^2 + 1520)]/2 = 19 making the two numbers 19 and 20.
19^2 + 20^2 = 361 + 400 = 761.

 

[lookagain]

... which reduces to x^2 + x - 380 = 0.

By means of the quadratic equation, \(\displaystyle > \ > \ >\)x = [-1+/-sqrt(1^2 + 1520)]/2 = 19\(\displaystyle < \ < \ <\)


\(\displaystyle Be \ careful \ here. \ \\) \(\displaystyle x \ = \ \frac{-1 \pm \sqrt{1^2 + 1520}}{2}, \ \ *from \ which* \ \ x \ = \ 19 \ \ is \ \ the \ only \ \ positive \ \ integer \ \ solution.\)


making the two numbers 19 and 20. 19^2 + \(\displaystyle >\ >\) 200^2 \(\displaystyle < \ <\) \(\displaystyle This \ is \ to \ be \ \ 20^2 \ \ instead. \ .\ . \ . \ . \ . \ (typo)\)

= 361 + 400 = 761.

 

[Denis]

Hmmm....divide 761 by 2; get 2 closest squares !!

19^2 = 361
761 / 2 = 380.5
20^2 = 400