Homework Help

How do a person solve and what rule is to be used when this problem x^2-5 is not a...

user profile pic

monique06 | (Level 3) Valedictorian

Posted May 10, 2013 at 11:15 AM via web

dislike 2 like

How do a person solve and what rule is to be used when this problem x^2-5 is not a perfect square?

2 Answers | Add Yours

user profile pic

embizze | High School Teacher | (Level 1) Educator Emeritus

Posted May 10, 2013 at 11:56 AM (Answer #1)

dislike 1 like

You are asked to solve `x^2-5=0` . (To "solve" a polynomial is to solve the equation where the polynomial is set equal to zero. Then you are looking for the solutions or roots of the polynomial.)

1. `x^2-5=0` Add 5 to both sides (Undo the subtraction)

`x^2=5`         Take the square root of both sides (Undo squaring)

`x=+-sqrt(5)`  which is the solution.

Note that there are two answers, because if `x=-sqrt(5)` , when you square it you will get 5.

Also, you may be asked to give exact answers (typical) which are `x=sqrt(5)"or"x=-sqrt(5)` . Or you may be asked to approximate the answers to a specified number of decimal digits. This is especially true in an applied course such as a science class.

2. You can use the quadratic formula: if `ax^2+bx+c=0` then `x=(-b+-sqrt(b^2-4ac))/(2a)` Given `x^2-5=0` we have a=1,b=0,c=-5 so:

`x=(0+-sqrt(0^2-4(1)(-5)))/(2(1))`

`x=+-sqrt(20)/2=(+-2sqrt(5))/2=+-sqrt(5)` as before.

3. With a graphing utility you can estimate the zeros (solutions) by finding the x-intercepts.

The grapher will be able to give approximations for the x intercepts.

user profile pic

oldnick | (Level 1) Valedictorian

Posted May 11, 2013 at 1:56 AM (Answer #2)

dislike 1 like

There is only two values of `x`  for  `x^2-5=y^2` with `y` integer.

Ideed :  if `x=y+n`  , `n`  integer:

`x^2=y^2+2yn+n^2`  and   `5= 2yn+n^2`  

Note, first, `n` is odd,  so or  `y` is even and  `x` odd, or the contrary.

On the other side we can compose `5=(x+y)(x-y)` 

Since `5` is prime: `x+y=1`  and `x-y=5` ``    or

`x+y=5`  and  `x-y=1`

so:     `x=3`   and  `y=-2`   in the first case,

or       `x=3`   and  `y=2`  in the second.

Now suppose , by absurd, exists another couple , `x_1;y_1` 

that:  `x_1^2-y_1^2=5`  different by the coulpes we've just found: `x_0;y_0`

So that:      `x_1^2-y_1^2=x_0^2-y_0^2`

            Since every other solution has to have sum or difference  between `x_1`  and `y_1`  dividing 5, so are to be equal solutions we've just found.

 

 

 

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes