Question

The equation x^5-3x^3+x^2-4=0 has one positive root. Verify by calculation that this root lies between 1 and 2. Can someone please solve this question?

Answer 1

A root of an equation is a value for the variable (in this case `x`) which makes the equation true. In other words, if we were to solve for `x`, then the solved value(s) would be the roots.

Usually when we talk about roots, it's with a function of `x`, like `y=x^5-3x^3+x^2-4`, and finding the roots means solving for `x` when `y` is 0.

If this function has a root between 1 and 2, then at some `x`-value between `x=1` and `x=2`, the equation will equal 0. Which also means that, at some point on one side of this root, the equation is positive, and at some point on the other side, it's negative.

Since we're trying to show that there's a root between 1 and 2, if we can show that the equation switches sign between these two values, we'll be done.

What is `y` when `x=1`?

`y=x^5-3x^3+x^2-4`
`color(white)y=(1)^5-3(1)^3+(1)^2-4`
`color(white)y=1-3+1-4`
`color(white)y=–5`
`color(white)y<0`

Now, what is `y` when `x=2`?

`y=x^5-3x^3+x^2-4`
`color(white)y=(2)^5-3(2)^3+(2)^2-4`
`color(white)y=32-3(8)+4-4`
`color(white)y=32-24`
`color(white)y=8`
`color(white)y>0`

We've shown that `y` is negative when `x=1`, and `y` is positive when `x=2`. So at some point between 1 and 2, there must a value for `x` which makes `y` equal to 0.

We've just used the Intermediate Value Theorem or (IVT). If you're not sure what that is, a quick description is that, if a continuous function is less than `c` when `x=a` and is greater than `c` when `x=b`, then at some point between `a` and `b`, the function must equal `c.`

Note:

The IVT is only applicable on continuous functions (or functions that are continuous on the interval of interest). Luckily, all polynomials in `x` are continuous everywhere, so that's why we can use the IVT here.