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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=+-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.
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.
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