Radical Equations And Problem Solving

So I'd have been checking my solutions for this question, even if they hadn't told me to.

I'll treat the two sides of this equation as two functions, and graph them, so I have some idea what to expect. This is for my own sense of confidence in my work.) I'll graph the two sides of the equation as: solution. It came from my squaring both sides of the original equation. I can see it in the squared functions and their graph: ("Extraneous", pronounced as "eck-STRAY-nee-uss", in this context means "mathematically correct, but not relevant or useful, as far as the original question is concerned".

And the best way to get rid of the 3 is to subtract 3 from the left-hand side.

And of course, if I do it on the left-hand side I also have to do it on the right-hand side.

The general process for isolation is, in a sense, undoing whatever had been done to the variable in the original equation.

For instance, suppose we are given the following linear equation: We can always check our solution to an equation by plugging that solution back into the original equation and making sure that it results in a true statement.For most of this lesson, we'll be working with square roots.For instance, this is a radical equation, because the variable is inside the square root: In general, we solve equations by isolating the variable; that is, we manipulate the equation to end up with the variable on one side of the "equals" sign, with a numerical value on the other side.When you do this-- when you square this, you get 5x plus 6. So we get x is equal to 15, but we need to make sure that this actually works for our original equation. And this is the principal root of 81 so it's positive 9.If you square the square root of 5x plus 6, you're going to get 5x plus 6. On the left-hand side, we have 5x and on the right-hand side, we have 75. We get x is equal to-- let's see, it's 15, right? Maybe this would have worked if this was the negative square root. So it's 3 plus 9 needs to be equal to 12, which is absolutely true. And this is where we actually lost some information because we would have also gotten this if we squared the negative square root of 5x plus 6. And now, this is just a straight up linear equation. So we need to make sure it actually works for the positive square root, for the principal square root. So we get 3 plus the principal square root of 5 times 15. And so that's why we have to be careful with the answers we get and actually make sure it works when the original equation was the principal square root. If the instructions don't tell you that you must check your answers, check them anyway.At the very least, compare your solution with a graph on your graphing calculator.(This is just one of many potential errors possible in mathematics.) To see how this works in our current context, let's look at a very simple radical equation: There is another way to look at this "no solution" difficulty.When we are solving an equation, we can view the process as trying to find where two lines intersect on a graph.


Comments Radical Equations And Problem Solving

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