Radical Equations And Problem Solving

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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".

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.

The left-hand side of the equation can be graphed as one curve, and the right-hand side of the equation can be graphed as another curve.

The solution to the original equation is the intersection of the two curves.But I'll check my solution at the end, anyway, because the instructions require it.First, I'll square both Because of this fact, my squaring of both sides of the equation will be an irreversible step.Otherwise, I would lose the ability to say that they're equal. And now, we can square both sides of this equation.And so the left-hand side right over here simplifies to the principal square root of 5x plus 6. So we could square the principal square root of 5x plus 6 and we can square 9. Or we get 3 plus square root of 75 plus 6 is 81 needs to be equal to 12.(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.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 term hasn't come up in your class yet, you should expect to hear it shortly.) By squaring both sides, I created an extra (and wrong) solution.Now I'll prove which solution is right by checking my answers.


Comments Radical Equations And Problem Solving

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