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Squaring negative square root binomials?
* signifies square root.
*2x+5+*2x-8=0 ( I'm doing problems such as these.)
The first step is to isolate each radical expression.
So, we have *2x+5= - *2x-8
Our next step is to square both sides of the equation.
(*2x+5= - *2x-8) ^2
Here's where some of my confusion comes into play beacuse I don't know what to do with the negative. Could someone give me the next step? My assumption would be to attach it to the two. So, I would square the *2x-8, then I would attach the negative to the two X. Is this correct or not?
2 Answers
- Anonymous4 years ago
sq. root is often valuable. to illustrate, enable x = -2 ---> 2(-2) + 3 = -a million 4(-2)^2 + 12(-2) + 9 = sixteen - 24 + 9 = a million =======================================... least complicated occasion of root(x^2) = absvalue(x) enable x = -3, x^2 = 9, root(x^2) = root(9) = 3 that's certainly the cost of -3. consistently works this way. once you spot root(x^2), think of absvalue(x)
