in-3-41-express-each-product-in-simplest-form-variables-in-the-radicand-with-an-even-index-are-no-35

Question

In $3-41$ , express each product in simplest form. Variables in the radicand with an even index are non-negative.
$$
(a+\sqrt{b})(a-\sqrt{b})
$$

Sarah Wharton · Accepted Answer

in this question, we&#x27;re going to use foil to solve a plus ruby times a minus ruby foil, meaning first outside inside last, so I&#x27;ll start with the 1st 8 times A. This is equal to a squared. Plus the outside comes next. A times negative root be. Don&#x27;t forget that negative sign, so it&#x27;s going to be a negative a route B. Now comes the inside root. Be times a so B plus a route to be finally, the last part. Positive ruby times and negative root be which will be a negative root. Be Times route. Be so it&#x27;s just negative be continuing to solve. We could notice that negative. A ruby and a ruby just cancel each other out, so our final answer will simply be a squared minus B.

Sarah Wharton · Answer

we&#x27;re solving for the quantity of three plus Route five a B cubed, the quantity squared. So I&#x27;m going to write this out to make it a little bit a little bit easier to solve all. Rewrite it as three plus Route five A. Be cute times three plus Route five A. Be cute, and now I&#x27;m going to use foil first outside, inside last to expand even further. It&#x27;s almost ply the first numbers together. I&#x27;ll read it on the second line as three times three is nine plus the outside numbers that would be three times Route five a B cubed. So three Route five A B cubed plus the inside numbers being multiplied together. That would be three. Route five a. B cubed again, and I&#x27;m actually going to make sure that this doesn&#x27;t look like a cubed root. It looks like a proper coefficients of three. Route five a B cubed. Do that on the other one to just to be clear. And then finally the last. So Route 5 80 cubed times, Route five A B cube. Anything times itself. If both of them are in a square root sign, it&#x27;ll just be the number that&#x27;s inside. So plus five A. Be cute. Now, finally, we need to combine like terms these two in the middle Since they have the same term inside the route, we just have to combine the coefficients, the three and the three. So this is going to be equal to nine plus three plus three is six. So six Route five A B cubed plus five a b cute. And that is our final answer.

M S · Answer

So we want to simplify this ex question. So the first thing I notice is that we have ah b squared over here. So this is too baby times to route a. So the square root of B squared so the square root in squared will cancel each other out. So we&#x27;ll have to route a ruby times three routing times. So nothing on Edison&#x27;s. We have this to a here and a Subaru a here. So two times he will give us four. And spirit of a town square Today will give us a and you live with the time to spirit be now because we have the same base here. We could simple far this further there is that I have for a be transferred of the

Sarah Wharton · Answer

in this problem, we&#x27;re going to use foil to solve nine plus route to be times one plus route to be starting with the F for first nine times, one is equal to nine, so that will be our first term in the expansion up. Next comes the outside numbers nine times route to be We&#x27;ll have plus nine route to be, plus the inside term one times route to be one route to be, plus the last route to be times route to be, which we can combine in one radical as route to be times to be. This could be simplified further as nine plus, recognizing that these two are like terms because they both have the route to be so. We nine plus 10 from the nine plus one route to be plus square root of two times two is four b times B is B squared. This four B squared can be broken down into root for times Route B squared. This in turn, is equal to two from the root. For times, be so really square root of for B Squared is just to be meaning that our final answer will be nine plus 10 route to be plus to be

Sarah Wharton · Answer

in this problem we&#x27;re solving for the square root of a over three times the square root of a squared over four. And for this problem, I&#x27;m gonna choose to simplify this term on the right first, because I noticed that both the top and the bottom are perfect squares. So this leaves me with the square root of a over three times the square root of a squared who were square root of four. This is equal to the square roots of a over three times A over to now in to deal with this term on the right. To do that, I&#x27;m going to similarly separate it into a top and bottom. So and I&#x27;m gonna put this term out in front just because I&#x27;m done dealing with it. So this is equal to a over two times the square root of a over the square root of three. Now, to get rid of this three on the bottom, when you do is multiply this term in particular by Route three over route three. Now, Route three over Route three is just a form of one. So we&#x27;re not fundamentally changing the equation at all were simply multiplying by one. But what that does is that kids get gets rid of the radical on the bottom here with those route threes. So this is going to be equal to a over to times Route three A. From combining the three and a there divided by route three times, Route three, which is just three now if I bring this on to another line and multiply altogether, this gives me a Times Route three a divided by two times three. So this is equal to a over six times Route three A and that is our final answer.

M S · Answer

So we have to square in 21 something spirit of 4/3. So one thing I noticed automatically is that if we split apart our second square root, so have a scary to four. On top of spirit of three, we can simplify our numerator. But I also noticed that three is a factor. 21 focus put apart our 21 We have skirted three times square to seven. So I&#x27;m screwed of four over square 23 Another square root of four is too. So we have three times. We&#x27;re seven times to over Route three. And since the rules applying, we can cancel this out with our three on the bottom, they&#x27;ll have to say I was a spirit of seven.

Problem

In $3-41$ , express each product in simplest form…

Problem 37 Hard Difficulty

In $3-41$ , express each product in simplest form. Variables in the radicand with an even index are non-negative.
$$
(a+\sqrt{b})(a-\sqrt{b})
$$

Answer

$=a^{2}-b$

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$=a^{2}-b$

Algebra 2 and Trigonometry

Ann Xavier GantertAlgebra 2 and Trigonometry

Ann Xavier Gantert

Algebra 2 and Trigonometry