Using the equation below as a model, fill in numbers in the place of a and b to create a rational equation that has an extraneous solution.
x+a/ax=b/x
Part 1. Show all work to solve for x in the equation and check the solution.
Part 2. Explain how to identify the extraneous solution and what it means.
Part 1:
To solve the given rational equation (x + a)/(ax) = b/x, we can start by cross-multiplying to eliminate the denominators.
First, let's multiply both sides of the equation by ax to get rid of the denominator on the left side:
(x + a) = (b/x) * ax
Expanding the right side of the equation:
x + a = b * (a/x)
Now, we can multiply both sides of the equation by x to eliminate the denominator on the right side:
x * (x + a) = b * a
Expanding the left side of the equation using the distributive property:
x^2 + ax = b * a
Rearranging the equation to get a quadratic equation in standard form:
x^2 + ax - b * a = 0
Now, we can solve the quadratic equation using the quadratic formula:
x = (-a ± √(a^2 - 4 * 1 * (-b * a))) / 2 * 1
Simplifying further:
x = (-a ± √(a^2 + 4ab)) / 2
We have found the solutions for x. Now, let's check if these solutions satisfy the original equation.
Part 2:
To determine if there are any extraneous solutions, we need to substitute each solution back into the original equation and verify if the equation holds true.
In our case, we substitute the values we found for x back into the original equation (x + a)/(ax) = b/x and simplify the equation.
Let's consider the "+" sign in the solution:
x = (-a + √(a^2 + 4ab)) / 2
Substituting x into the original equation gives us:
((-a + √(a^2 + 4ab)) + a) / (a((-a + √(a^2 + 4ab)) / 2)) = b / ((-a + √(a^2 + 4ab)) / 2)
Simplifying the equation further:
((-a + √(a^2 + 4ab)) + a) / (-a + √(a^2 + 4ab)) = b
Next, let's consider the "-" sign in the solution:
x = (-a - √(a^2 + 4ab)) / 2
Substituting x into the original equation gives us:
(((-a) - √(a^2 + 4ab)) + a) / (a(((-a) - √(a^2 + 4ab)) / 2)) = b / ((-a - √(a^2 + 4ab)) / 2)
Simplifying the equation further:
(((-a) - √(a^2 + 4ab)) + a) / (-a - √(a^2 + 4ab)) = b
Now, if we simplify both equations, we may find that one or both sides of the equation do not remain equal. In such a case, we have an extraneous solution.
Therefore, to identify the extraneous solution, we compare both sides of the equation in each case. If any solution leads to an inequality, for example, by resulting in a division by zero or an equation that does not hold, that particular solution is an extraneous solution.
In summary, to identify an extraneous solution in a rational equation, you should substitute the solution(s) back into the original equation and verify that both sides of the equation remain equal. If any solution results in an inequality or an equation that does not hold, it is an extraneous solution.
Did you mean it the way you typed it, then by the order of operations....
x + 1x= b/x, since a/a = 1
2x = b/x
2x^2 = b
x^2 = √(b/2)
for extraneous roots, a can be any non-zero number, and b < 0
If you meant:
(x+a)/(ax) = b/x
x^2 + ax = abx
x^2 + ax - abx = 0
x^2 + x(a - ab) = 0
for extraneous roots the discriminant < 0
(a-ab)^2 - 4(1)(0) < 0
(a - ab)^2 < 0
but any square is ≥ 0, so no combination of a and b will produce extraneous roots
if you meant:
((x+a)/a)(x) = b/x
...
you try it