Question

the equation + = - has the solution set {}. : Explain why the equation ( - ) + = ( -) + also has the solution set of {} : when we square both sides of the original equation, we get the following new equation: ( + )² = ( - )². Show that {} is still a solution to the new equation : show that {} is also a solution to the new equation. (6x +8)² +(10x - 8)² , but is not a solution the original equation, 6x +8 = 10x - 8. : write a sentence that describes how the solution set to an equation may change when both sides of the equation are squared.

Answer 1

The solution of 6x+8=10x-8 is x=4:

`\begin{gathered} 6x+8=10x-8 \\ 6x-10x=-8-8 \\ -4x=-16 \\ x=4 \end{gathered}`

Part A:

We can write:

`\begin{gathered} (6x+8)+2=(10x-8)+2 \\ 6x+8=10x-8 \\ x=4 \end{gathered}`

The equation in the second step is the same as the first equation, that has the solution set x=4.

Part B:

We have the first equation squared:

`\begin{gathered} (6x+8)^2=(10x-8)^2 \\ \sqrt[]{(6x+8)^2}=\sqrt[]{(10x-8)^2} \\ |6x+8|=|10x-8| \\ 6x+8=10x-8 \\ x=4 \end{gathered}`

Then we have the same equation that we started from.

Part C:

We have a new equation and we have to prove that x=0 is a solution:

`\begin{gathered} (6x+8)^2=(10x-8)^2 \\ (6\cdot0+8)^2=(10\cdot0-8)^2 \\ 8^2=(-8)^2 \\ 64=64\longrightarrow\text{True} \end{gathered}`

Then, x=0 is a solution.

But if we apply the same to the first equation we get:

`\begin{gathered} 6x+8=10x-8 \\ 6\cdot0+8=10\cdot0-8 \\ 8=-8\longrightarrow\text{False (x=0 is not a solution)} \end{gathered}`

Part D:

When both sides of the equation are squared, both sides will have a positive sign. This does not necesarily happen when the sides are not squared.

For example, the equation we have been working for, when squared, is equal to:

`|6x+8|=|10x-8|`

That is why x=0 is a solution for the squared equation but not for the original equation, when both sides end with different sign.