a To solve an equation we should first gather all of the variable terms on one side and all of the constant terms on the other side using the Properties of Equality. In this case,
we need to start by subtracting 23456 from both sides of the equation. Use a calculator if needed.
b To solve the given equation, we should first simplify the fractions containing the variable x. To do so, let's start by multiplying the equation by the least common denominator (LCD). Then we will use the Properties of Equality to solve for x.
ax^2+ bx+ c=0 ⇕ x=- b± sqrt(b^2-4 a c)/2 a
We first need to identify the values of a, b, and c.
5x^2-6x+1=0 ⇕ 5x^2+( - 6)x+ 1=0
We see that a= 5, b= - 6, and c= 1. Let's substitute these values into the Quadratic Formula.
Using the Quadratic Formula, we found that the solutions of the given equation are x= 6± 410.
x=6± 4/10
x_1=6+4/10
x_2=6-4/10
x_1=10/10
x_2=2/10
x_1= 1
x_2= 1/5
Therefore, the solutions are x_1=1 and x_2= 15.
d To solve the given equation, we will factor out the greatest common factor. In our case, x is the greatest common factor.
x^3-3x^2+2x = 0
⇕
x( x^2-3x+2 ) =0
We have rewritten the left-hand side as a product of two factors. Now, we will apply the Zero Product Property to solve the equation.
x( x^2-3x+2 ) =0
⇕
lcx=0 & (I) x^2-3x+2=0 & (II)From Equation (I) we found that one solution is x=0. To find the other solutions we will solve Equation (II). Note that this is a quadratic equation. Thus, we will use the Quadratic Formula.
ax^2+bx+c=0
⇕
x=- b±sqrt(b^2-4ac)/2a
To do so, we first need to identify a, b, and c.
x^2-3x+2=0
⇕
1x^2+( - 3)x+ 2=0
We see that a= 1, b= - 3, and c= 2. Let's substitute these values into the formula and solve for x.
Using the Quadratic Formula, we found that the solutions of the given equation are x= 3± 12.
x=3± 1/2
x_1=3+1/2
x_2=3-1/2
x_1=4/2
x_2=2/2
x_1= 2
x_2= 1
Therefore, the solutions to the quadratic equation are x_1=2 and x_2= 1. These solutions are also solutions for the given equation.
Solutions
x=0, x=1, x=2
