In Exercises 1–38, solve each radical equation._____ _____√6x + 2...

Question

In Exercises 1–38, solve each radical equation._____ _____√6x + 2 = √5x + 3

Accepted Answer

Welcome back. I am so glad you're here. We are told in the given radical equation to solve for X. Our given radical equation is on the left, the square root of 13 X plus eight. And that's equal to, on the right, the square root of 12 X plus 23. And our answer choices are answer choice A, the solution set negative 15 answer choice B the solution set 15 answer choice C, the solution set eight and answer choice D the solution set 23. All right. Typically when we're solving for X, we want to get X by itself on one side of the equal sign and all the numbers to the other side of the equal sign. This time we have two XS one on each side of the equal sign and they're both trapped underneath radicals. So first, we need to free these XS from their radicals. And when we take the square root of something, the way to undo that square root is to square it. And it's very important when you take one side and square it, you need to square the other side as well. It just so happens that doing that will free both of these Xes from their radicals simultaneously. Yay. So on the left, taking the square root of 13 X plus eight squared will leave us with just 13 X plus eight. And that equals on the right, taking the square root of 12 X plus squared will leave us with just 12 X plus 23. And now is where we get the X by itself on one side of the equal sign and all the numbers to the other side, we need to move the XS to the same side. It could be either side. But what I like to do when possible is to make it so that the number in front of the X, it's coefficient is positive. So I'm going to subtract 12 X from both sides of the equal sign. On the right hand side, the positive 12 X minus 12 X will equal zero. So that X that those 12 Xes have moved over to the left hand side of the equal sign. I am also going to subtract eight from both sides of the equal sign because on the left, that positive eight minus eight will be zero. And then we won't have those eights on the same side as our XS. So now simplifying 13 X minus 12 X, on the left hand side of the equal sign leaves us with one X or just X and that equals on the right hand side 23 minus eight, which is a positive 15. And we have done it, we have solved for X, we got it by itself and we look at our answer choices for X equals 15 and that matches with answer choice. B well done. We'll catch you on the next one.

In Exercises 1–38, solve each radical equation._ _√6x + 2 = √5x + 3

Key Concepts

Radical Equations

Isolating the Radical

Extraneous Solutions

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