Solve the equation x3 - 3x2 - 4x + 12 = 0, etc

[marshall1432]

solving equations:

1. Solve the equation x3 - 3x2 - 4x + 12 = 0.

i took the 12 to the other side of the equals bar and got: x3-3x2-4x=-12

2. Solve the equation (2x - 1)1/2 = 7.

i multiplied both numbers in paranthesis by 1/2 and then took the number to the other side of the fraction bqar:

(x-1/2)=7
(x)=7 1/2

3. Solve the equation x6 - 1 = 0. Find only the real number solutions.

I know it has something to do with finding x5,x4,x3,x2,x1 but i do not understand how to find those values.

4. Solve the absolute value equation | 6x + 2 | = 5.

i took the 2 to the other side of the equals bar and got:

|6x|=3
x=1/2

5. Solve the equation 2x4 - 3x3 - 2x2 = 0.

Same as 3, i have to find the x value but again, how do i get it?

thanks guys for all your help!

 

[soroban]

Re: solving equations

Hello, marshall1432!

Evidently you're not familiar with the first method for solving polynomial equations.
. . (1) Factor the polynomial (if possible).
. . (2) Set each factor equal to zero and solve each equation.

1. Solve: \(\displaystyle \,x^3\,-\,3x^2\,-\,4x\,+\,12\:=\:0\)

Factor: \(\displaystyle \:x^2(x\,-\,3)\,-\,4(x\,-\,3)\:=\:0\)

Factor: \(\displaystyle \x\,-\,3)(x^2\,-\,4)\:=\:0\)

Factor: \(\displaystyle \x\,-\,3)(x\,-\,2)(x\,+\,2)\:=\:0\)


Set each factor equal to zero and solve for \(\displaystyle x.\)

. . \(\displaystyle \begin{array}{ccc}x\,-\,3\:=\:0 & \;\Rightarrow\; & \fbox{x\,=\,3} \\
x\,-\,2\:=\:0 & \;\Rightarrow\; & \fbox{x\,=\,2} \\
x\,+\,2\:=\:0 & \;\Rightarrow\; & \fbox{x\,=\,-2} \end{array}\)

2. Solve: \(\displaystyle \,(2x\,-\,1)^{\frac{1}{2}} \:=\:7\)

That's not a "times one-half" . . . it's a one-half power (a square root).

Square both sides: \(\displaystyle \left[(2x\,-\,1)^{\frac{1}{2}}\right]^2 \:=\:7^2\)

. . and we have: \(\displaystyle \:2x\,-\,1\:=\:49\)

Can you finish it now?

3. Solve: \(\displaystyle \,x^6\,-\,1\:=\:0\;\;\)Find only the real number solutions.

This takes some fancy factoring . . .

We have: \(\displaystyle \,(x^3)^2\,-\,1^2\) . . . a difference of squares

Factor: \(\displaystyle \x^3\,-\,1)(x^3\,+\,1)\:=\:0\) . . . a difference of cubes and a sum of cubes

Factor: \(\displaystyle \x\,-\,1)(x^2\,+\,x\,+\,1)(x\,+\,1)(x^2\,-\,x\,+\,1)\:=\:0\)


Set each factor equal to zero and solve . . .

. . \(\displaystyle \begin{array}{cccc}x\,-\,1\:=\:0 & \;\Rightarrow\; & \fbox{x\,=\,1} \\
x^2\,+\,x\,+\,1\:=\:0 & \Rightarrow & \text{no real roots} \\
x\,+\,1\:=\:0 & \;\Rightarrow\; & \fbox{x\,=\,-1} \\
x^2\,-\,x\,+\,1\:=\:0 & \Rightarrow & \text{no real roots} \end{array}\)

4. Solve: \(\displaystyle \,|6x\,+\,2|\:=\:5\)

Evidently you don't understand absolute values.

This is equavalent to two equations:
. . \(\displaystyle 6x\,+\,2\:=\:5\;\;\Rightarrow\;\;6x\,=\,3\;\;\Rightarrow\;\;\fbox{x\,=\,\frac{1}{2}}\)
. . \(\displaystyle 6x\,+\,2\:=\:-5\;\;\Rightarrow\;\;6x\,=\,-7\;\;\Rightarrow\;\;\fbox{x\,=\,-\frac{7}{6}}\)

5. Solve: \(\displaystyle \,2x^4\,-\,3x^3\,-\,2x^2\:=\:0.\)

"Moving the constant over" is not the first step.
Did someone tell you to do that ... every time?

Factor: \(\displaystyle \;x^2(2x^2\,-\,3x\,-\,2)\:=\:0\)

Factor: \(\displaystyle \:x^2(x\,-\,2)(2x\,+\,1)\:=\:0\)

. . \(\displaystyle x^2\:=\:0\;\;\Rightarrow\;\;\fbox{x\,=\,0}\)
. . \(\displaystyle x\,-\,2\:=\:0\;\;\Rightarrow\;\;\fbox{x\,=\,2}\)
. . \(\displaystyle 2x\,+\,1\:=\:0\;\;\Rightarrow\;\;\fbox{x\,=\,-\frac{1}{2}}\)

 

[marshall1432]

thanks!

thank you so much! now i can take what you showed me and go back and work out some extra problems myself. i now know how to factor these types of problems, thank you so much for your help!

marshall