An absolute value measures an expression's distance from a midpoint on a number line.
|x-3| = 2
This equation means that the distance is 2, either in the positive direction or the negative direction.
|x-3|= 2 ⇒ lx-3= 2 x-3= - 2
To find the solutions to the absolute value equation, we need to solve both of these cases for x.
Both 5 and 1 are solutions to the absolute value equation.
b
To solve the given quadratic equation, we can isolate the variable expression raised to the power of 2 and take the square roots. Remember to consider the positive and negative solutions.
We found that x=± 2 + 2. We will now find both solutions by using the positive and negative signs.
x=± 2 + 2
x=2 + 2
x=-2 + 2
x=4
x=0
There are two solutions to the given equation, x=4 and x=0.
c
To solve equations with a variable expression inside a radical, we first want to make sure the radical is isolated. Then we can raise both sides of the equation to a power equal to the index of the radical. Note that the square root in the given equation is already isolated.
We first need to identify the values of a, b, and c.
x^2-5x-14=0 ⇕ 1x^2+( - 5)x+( -14)=0
We see that a= 1, b= - 5, and c= -14. Let's substitute these values into the Quadratic Formula.
Using the Quadratic Formula, we found that the solutions of the given equation are x= 5± 92.
x=5± 9/2
x_1=5+9/2
x_2=5-9/2
x_1=14/2
x_2=-4/2
x_1= 7
x_2= -2
The solutions are x_1=7 and x_2= -2. Finally, we will check for extraneous solutions by substituting both x= 7 and x= -2 into the original equation. Let's start with x= 7.
This time we obtained a false statement. Therefore, x= -2 is not a solution to the original equation.
d
An absolute value measures an expression's distance from a midpoint on a number line.
|2x+5|= 3x+4
This equation means that the distance is 3x+4, either in the positive direction or the negative direction.
|2x+5|= 3x+4
⇓
l2x+5= (3x+4) 2x+5= -(3x+4)
To find the solutions to the absolute value equation, we need to solve both of these cases for n.
Both x=1 and x=- 95 are solutions to the equation. Finally, we will check for extraneous solutions by substituting both x=1 and x=- 95 into the original equation. Let's start with x=1.
