Solving equations

[rachelmaddie]

In one sense, the MECHANICS of algebra are about moving step by step from a complex equation to a very simple equation.

[MATH]6 + \sqrt[5]{249 - 2x} = 7.[/MATH] Original complex equation

[MATH]\sqrt{249 - 2x} = 1.[/MATH] Simpler. Root is isolated by itself.

[MATH](\sqrt[5]{249 - 2x})^5 = 1^5.[/MATH] Raise both sides of previous equation to 5th power

[MATH]249 - 2x = 1.[/MATH] Simpler now for sure because the root was eliminated.

[MATH]249 - 1 = 2x.[/MATH] Unknown is isolated.

[MATH]2x = 248.[/MATH] Very simple equation.

[MATH]x = 124.[/MATH]
Solving an algebraic equation involves taking steps to arrive at the super-simple equation x = some numeral.

Do I need to make any more changes?

 

[JeffM]

No. what you did was great.

I was merely answering your question of "which equation" and trying to make a point about the mechanics of finding numerical solutions to algebraic equations. I personally have always found it helpful to realize that those mechanics aim at arriving at the simplicity of x = some numeral. All the techniques that you learn are means to achieve that extreme simplicity.

 

[HallsofIvy]

Be careful! When multiplying both sides of an equation by something that involves the unknown, or when raising both sides or an equation to a power you may introduce "new roots" that satisfy the new equation but not the original equation.

As an obvious example, the equation x= 5 has the single root, 5. If you multiply both sides of the equation by x you get \(\displaystyle x^2= 5x\) of \(\displaystyle x^2- 5x= 0\) which has roots 0 and 5. Similarly squaring both sides gives \(\displaystyle x^2= 25\) which has roots 5 and -5.

Always check your answers in the original equation.

 

[rachelmaddie]

Be careful! When multiplying both sides of an equation by something that involves the unknown, or when raising both sides or an equation to a power you may introduce "new roots" that satisfy the new equation but not the original equation.

As an obvious example, the equation x= 5 has the single root, 5. If you multiply both sides of the equation by x you get \(\displaystyle x^2= 5x\) of \(\displaystyle x^2- 5x= 0\) which has roots 0 and 5. Similarly squaring both sides gives \(\displaystyle x^2= 25\) which has roots 5 and -5.

Always check your answers in the original equation.

Are you talking about my original work or the revisions made?

 

[HallsofIvy]

In your original post you raised both sides of the equation to the 5th power, arriving at a linear equation to get x= 124 but did not check to see if that value of x actually satisfies the original equation. That is very dangerous. In changing from one equation to another it is possible to introduce "spurious solutions" that satisfy one equation but not the other.