Radical Equations And Problem Solving

Radical Equations And Problem Solving-71
On the left-hand side of this equation, I have a square root. On the right-hand side, I've got a positive number.Since both sides are known positive, squaring won't introduce extraneous solutions. If the instructions don't tell you that you must check your answers, check them anyway.

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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.

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.

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".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.My check is done by plugging the proposed solution into both the left-hand side (LHS) and right-hand side (RHS) of the original equation, and confirming that each simplifies to the same value (or else showing that the solution isn't any good): Even if the instructions hadn't told me to check my answers, clearly I needed to.And I needed to do that check with algebra, not with the picture.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.But you have to be very careful there because when you square radical signs you actually lose the information that you were taking the principal square root. So the first thing I want to do is I want to isolate this on one side of the equation.Not the negative square root or not the plus or minus square root. And so when we get our final answer, we do have to check and make sure that it gels with taking the principal square root. And the best way to isolate that is to get rid of this 3.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.

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