Write all the terms on one side of the equation and simplify as much as possible.
C
c
Rearrange the radical equation so that the radical expression is isolated. Then, raise both sides of the equation to a power equal to the index of the radicals.
3/x+2/x+1=5
Let's begin by moving the variables out of the denominators. To do so, we should multiply both sides of the equation by the least common denominator (LCD). This is the product of the factors in the denominators.
3/x+2/x+1=5Therefore, let's multiply both sides of the equation by x (x+1).
Note that we have a quadratic equation now. Since both sides are non-negative, we can take the square root from each side of the equation. Keep in mind that we need to consider the positive and negative solutions.
Finally, we will check if either x=- sqrt(35) or x=sqrt(35) is an extraneous solution. To do this we need to substitute them for x in the original equation. Let's start with x=- 35.
Neither of our solutions is an extraneous solution. Therefore, the solutions to the given equation are x=±sqrt(35).
b
We want to solve the following equation.
x^2+6x+9=2x^2+3x+5
Since it is a quadratic equation, we can solve it using the Quadratic Formula.
ax^2+ bx+ c=0 ⇕ x=- b± sqrt(b^2-4 a c)/2 aBefore we do that we should write all the terms on one side of the equation and simplify as much as possible.
Now we can identify the values of a, b, and c.
0=x^2-3x-4 ⇕ 1x^2+( - 3)x+( - 4)=0
We see that a= 1, b= - 3, and c= - 4. Let's substitute these values into the Quadratic Formula.
The solutions for this equation are x= 3±52. Let's separate them into the positive and negative cases.
x=3±5/2
x_1=3+5/2
x_2=3-5/2
x_1=8/2
x_2=-2/2
x_1=4
x_2=-1
Using the Quadratic Formula, we found that the solutions of the given equation are x_1=4 and x_2=-1. Let's check if any of them is an extraneous solution by substituting the values into the original equation.
We found that x=-1 creates a true statement as well. Therefore, the solutions to the equation are 4 and -1.
c
To solve equations with a variable expression inside a radical, we will first rearrange the radical equation so that the radical expression is isolated. Then we can raise both sides of the equation to a power equal to the index of the radicals. Let's try to solve our equation using this method!
Since adding or subtracting zero does not change the value of a number, the numerator will simplify to 8. Therefore, we will get only one solution.
x= 8/2 ⇔ x= 4
Using the Quadratic Formula, we found that the solution of the given equation is x=4. Let's substitute it into the original equation to check if it is extraneous.
