Show that the equation x^3+ x^2-3=0 has no rational roots, but that it does have an irrational root between x=1 and x=2
if x^3 + x^2 - 3 = 0
has any rational roots then they must be one of
-3 or +3
that is there should be a factor of either (x+3) or (x-3)
Using either long algebraic division, or synthetic division, it can be shown that neither divides into it evenly.
However,
f(1) = 1 + 1 - 3 = -1
f(2) = 8 + 4 - 3 = 9
Since the curve it continuous, and at x = 1 it is below the x-axis, and at x = 2 it lies above the x-axis, for some value of x between 1 and 2 it must have crossed the x-axis, which would be the root.
A quick sketch will show that it can cross only once,
http://www.wolframalpha.com/input/?i=x%5E3%2B+x%5E2+-+3+%3D+0
Sure, let's have some math fun!
To determine whether the equation x^3 + x^2 - 3 = 0 has rational roots or not, we can use the Rational Root Theorem. According to the theorem, if a rational root p/q (where p and q are integers) exists, it must satisfy the conditions: p divides the constant term (-3), and q divides the leading coefficient (1).
Let's list all the possible rational roots: ±1, ±3. To our disappointment, none of these values satisfy the equation x^3 + x^2 - 3 = 0. So, there are no rational roots. Sorry, math can be downright irrational sometimes!
Now, let's move on to the fun part! We need to show that there exists an irrational root between x = 1 and x = 2. To do this, one way is to use the Intermediate Value Theorem.
The Intermediate Value Theorem states that if a function is continuous on an interval [a, b] and takes on two values of opposite sign (meaning, one positive and one negative) at the endpoints, then there must exist at least one root within the interval.
Now, let's evaluate the function at x = 1 and x = 2:
f(1) = 1^3 + 1^2 - 3 = -1
f(2) = 2^3 + 2^2 - 3 = 9
Since f(1) = -1 and f(2) = 9, f(x) changes sign between x = 1 and x = 2. Therefore, by the Intermediate Value Theorem, we can conclude that there exists at least one root between x = 1 and x = 2.
So, let's celebrate the existence of this sneaky irrational root! It's like finding a clown hiding in a box of calculators. Unpredictable and full of surprises!
To show that the equation x^3 + x^2 - 3 = 0 has no rational roots, we can use the Rational Root Theorem. According to the theorem, any rational root of the equation must be of the form p/q, where p is a factor of the constant term (-3) and q is a factor of the leading coefficient (1).
First, let's find the factors of -3: ±1, ±3.
Next, let's find the factors of 1: ±1.
Therefore, the possible rational roots are: ±1/1, ±3/1, and none of these roots satisfy the equation.
Now, let's show that the equation does have an irrational root between x = 1 and x = 2. We can use the Intermediate Value Theorem.
Consider f(x) = x^3 + x^2 - 3.
f(1) = (1)^3 + (1)^2 - 3 = 1 + 1 - 3 = -1.
f(2) = (2)^3 + (2)^2 - 3 = 8 + 4 - 3 = 9.
Since f(1) = -1 and f(2) = 9, and f(x) is a continuous function, according to the Intermediate Value Theorem, there must be a root of f(x) = 0 between x = 1 and x = 2. Since the equation has no rational roots, this root must be an irrational number.
To show that the equation has no rational roots, we can use the Rational Root Theorem. The Rational Root Theorem states that if a polynomial has a rational root, it must be of the form p/q, where p is a factor of the constant term (-3 in this case) and q is a factor of the leading coefficient (1 in this case). So the possible rational roots for this equation are ±1, ±3.
To check if these values are roots, we substitute them into the equation:
For x = -1: (-1)^3 + (-1)^2 - 3 = -1 + 1 - 3 = -3, which is not zero.
For x = 1: (1)^3 + (1)^2 - 3 = 1 + 1 - 3 = -1, which is not zero.
For x = -3: (-3)^3 + (-3)^2 - 3 = -27 + 9 - 3 = -21, which is not zero.
For x = 3: (3)^3 + (3)^2 - 3 = 27 + 9 - 3 = 33, which is not zero.
Since none of the possible rational roots are actual roots of the equation, we can conclude that the equation has no rational roots.
To show that the equation has an irrational root between x = 1 and x = 2, we can use the Intermediate Value Theorem. The Intermediate Value Theorem states that if a continuous function takes on values of opposite signs at two points, then there exists a root between those two points.
Let's evaluate the function at x = 1 and x = 2:
For x = 1: (1)^3 + (1)^2 - 3 = -1
For x = 2: (2)^3 + (2)^2 - 3 = 11
Since the function takes on values of opposite signs at x = 1 and x = 2 (-1 and 11, respectively), we can conclude that there exists at least one root between x = 1 and x = 2.
Therefore, the equation x^3 + x^2 - 3 = 0 has no rational roots, but it does have an irrational root between x = 1 and x = 2.