in-9-26-solve-each-quadratic-equation-by-completing-the-square-express-the-answer-in-simplest-radi-4

Question

In $9-26,$ solve each quadratic equation by completing the square. Express the answer in simplest radical form.
$$
x^{2}+2 x-5=0
$$

Aditya Balasubramanian · Accepted Answer

kids. The right theory of the expression expert plus two X minus five is equal to zero and we want to solve for X You seen competing square. And if you look at this two x in the middle right here, you will see that you have. Uh, if you have a square root of that, you would have to use one X plus one ex. So in order to get that two exes, that second term. So let&#x27;s pretend you use X plus one squared. Great. Because X squared plus one plus one ex again would create that middle term right there. But X plus one squared also would create a plus one at the end. So in order to account for that, let&#x27;s review right originally given problem, that&#x27;s your right as X squared plus two x, that&#x27;s right. The plus one here so you can isolate the X plus one squared. And in order for this to work, you have the right side would have to be equal 26 right, cause you do one minus six and you bring over the six other side. You creating negative five on this left side right here. So my little itself, right? You get X plus one squared is equal to six square both sides so that we can solve for X Big X plus one equals plus or minus root. 62 Solutions Because we&#x27;re solving for a root frost given square, this is the one right here in case I wasn&#x27;t clear before, Um, finally X equals plus or minus Route six. Bring over that one minus one, and that is solving for X. It&#x27;s most simplest form.

Foster Wisusik · Answer

Okay, This question would like us to solve this quadratic equation. So in order to do this, we&#x27;re going to use the method of completing the square. Will we turn the general quadratic into a special form that makes it easier to solve equations with? And to do this will need to find two constants. One is D, which is equal to be over to a and the other is E, which is C minus B squared over for a So all we have to do is find d and E for our quadratic so d is equal to be over to a which in our case, bees negative too in a is one. So we just have negative one for B and then for E. We have C minus B squared over four a and C is negative two, and B squared is four divided by four. So we have negative two minus one. So negative three for E. So this turns our quadratic into a is one x plus d squared plus e. And now this equation is much easier to solve than the one we started with. So now let&#x27;s add three to both sides to get X minus one. Quantity squared is equal to three. So then we get X minus. One is equal to plus or minus the square to three. So then we have two different scenarios. So first we have the positive route and then we have the negative root. So for our first case, adding wonderful sides, we get X equals one plus Route three. And then for the other route we get X equals one minus the squared of three. And these are our solutions to the quadratic.

Foster Wisusik · Answer

Okay, This question wants us to solve the following quadratic. So we&#x27;re going to use the method of completing the square which turns X squared plus B x plus e into Vertex for a Times X plus d squared plus e. And we said D is defined to be be over to a and e is defined as C minus B squared over for a So let&#x27;s calculate these constants for our quadratic and Palladian. So we said that d is be over to a which in our example bees negative six days, one so d is negative three and then e is to minus B squared, which is 36 over for a So we get tu minus nine, which is negative seven. So are quadratic can now be it written in this reduced form as X minus three quantity squared minus seven is equal to zero, cause we just plugged into our complete the square form after we found D and E. So now we&#x27;re gonna add seven double sides to get X minus. Three squared is equal to seven or X minus three is equal to plus or minus the square root of seven. So now we&#x27;re gonna get to solutions we get one branch for X minus three equals positive Square seven and one branch for X minus three equals negative Route seven. So let&#x27;s just add three double sides to get our solutions of X equals three plus Route seven and X equals three minus Route seven and those air answers.

Foster Wisusik · Answer

Okay, This question would like us to solve the following quadratic. So to start, we&#x27;re gonna use to complete the square method which turns a general quadratic X squared plus B x plus e into a form that&#x27;s easier to solve. Equations with a times X plus d squared plus e for some constant. Do you? A. So we said that D is equal to be over to A and E is equal to C minus B squared over four a. So now let&#x27;s find these d any values for our quadratic. So for D, that&#x27;s just be over to a which is negative for over two or negative, too. And then for E. We have C minus B squared over for a and that&#x27;s one minus the square to 16 and four times eight&#x27;s for, so we get one minus for more negative three. So now are quadratic turns into X minus two. All squared minus three equals zero, and that&#x27;s a much easier equation to solve. So now we get X minus two. Quantity squared is equal to three. When we add three of all sides, and then if we take the square root, we get X minus two is equal to plus or minus the square root of three. So we know that we have two solutions. We have X minus three. Person X minus two equals the positive squared of three and X minus two equals the negative squared of three. So adding to teach equation gives us one solution of two plus squared of three and another solution of two minus squared of three. And these are our answers.

Foster Wisusik · Answer

Okay, This question would like us to solve the following quadratic by completing the square. So to start, we&#x27;re just going to use the fact that a X squared plus B X plus e could be transformed into a Times X plus d squared plus e, where D is be over to a and e is C minus. B squared over for a So now let&#x27;s find these values for our quadratic So d is equal Toby over to a, which is 12 over two times two. So we have 12/4, which is just three and then for E. We have C minus B squared over four A, which is three minus 12 squared is 1 44 divided by for a, which is eight. So we get three minus 1 44 over AIDS, and this simplifies to three minus 18 which is negative. 15. So now are quadratic can be written as a Times X plus D squared plus e equals zero. So now we can go ahead and solve this equation, so adding 15 to each side we get two times X plus three is equal to 15. But keep this square on and then X plus three quantity squared is equal to 15 half&#x27;s or X plus three is equal to the square root of 15 halves. And then, since we&#x27;re taking a square root, this could be positive or negative. So the next thing we can dio is find the solutions on each branch. So we have X Plus three equals positive squared, 15/2, and then we have X Plus three equals negative Route 15 over to. So solving this we get X equals negative three, plus Route 15 over to and X equals negative three, minus Route 15 over to. But this might not be a convenient form for a final answer, so we can actually simplify this by getting rid of the radical in the denominator, because we really don&#x27;t like those. So let&#x27;s multiply both of the used by route to over to route you over to to get rid of the radical in the denominator. So we get X equals negative three, plus Route 30 over to and X equals negative three minus route 30 over two. What if you really want to combine them into one fraction, you can get that three in the common denominator of two by multiplying top and bottom by to to get negative six plus Route 30 all over, too. And X equals negative six minus Route 30 all over, too. And these are our final answers.

Aditya Balasubramanian · Answer

So I hear we have a function to X squared minus six X plus three this equal to zero in order to use completing the square. Let&#x27;s start off by factoring this out right, you can&#x27;t. It&#x27;s hard to do this if you have a, um, a co option of two in the X squared. So it&#x27;s factor this two out. And if you factored to Alec, you can also factor two out of the six negative six x so it will become two times X squared, minus three X. Keep the three out of it for now. Thank you, Thank you. Now, if you take a look at this native free X, this is technically the second term of and expressed form or expanded form of apologetic expression, and you could use the roots Negative three halves number three house times two people, native three. So let&#x27;s get this room handy. Red pen over here. Let&#x27;s pretend we use X minus three halves squared. So what would occur if we would get two times X squared? I&#x27;m gonna write this expense. We have squared and expanded form. Just concede a little more clearly. X squared minus three X and then three have some students. We need this final term that is equal to nine forts, correct. So it&#x27;s at night worths, all right, red pen. And we also have to subtract 9/4. But since we don&#x27;t want to interfere with this nice little expression, you got going on the parentheses because it&#x27;s it&#x27;ll be clean to just simplify like that. Well, multiply that negative nine forts by two. So will become negative nine haves and leave it out of there so you can actually simplify this X squared minus three hacks plus 9/4 R X squared minus three X plus nine for it&#x27;s a little Easier. So let&#x27;s do that. Let&#x27;s it&#x27;s untracked nine haves and keep it on the outside and then once again at that three from before and we would get that school gets two times X minus three halves squared not to add the equal zero. Originally, colleges that you get two times X minus three zero X rays. Three Scored the expense we have squared. That&#x27;s not very clear, but remember, that&#x27;s just a negative two halves right there. Tim&#x27;s excited. You have squared, um, native nine halves, plus the results of six hands so it would be minus three. Have negative naps plus six absence. Major three has men. That was a zero just lining up the science of clarity. And then what we would that have to do is solve for X right. You want to find X in its simplest form. So we would get two plums. X ministry have squared. Been that native three halves over. You have the right side. Then you have x three have squared. It was three halves divided by to correct. So three warrants then square both sides. You get X equals Dr Emmett experience. They have equals. We have not moved three halves already, and we have first you if you&#x27;d say X minus three has because plus or minus root of 3/4 and therefore X equals closer minus root 3/4 plus three hands and that is it. That is X we&#x27;re solving for X that is using competing square. And initially we had a lot of factoring you do, and this is a little trickier than if you had a coefficient of one, but that whenever you see a coefficient that it&#x27;s not one you&#x27;d have to know to back here. Now what&#x27;s it

Problem

In $9-26,$ solve each quadratic equation by compl…

Problem 18 Medium Difficulty

In $9-26,$ solve each quadratic equation by completing the square. Express the answer in simplest radical form.
$$
x^{2}+2 x-5=0
$$

Answer

$-1+\sqrt{6}$ and $-1-\sqrt{6}$

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Ann Xavier Gantert

Algebra 2 and Trigonometry

Grace H.

Anna Marie V.

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Absolute Value - Example 1

Absolute Value - Example 2

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$-1+\sqrt{6}$ and $-1-\sqrt{6}$

Algebra 2 and Trigonometry

Grace H.

Anna Marie V.

Heather Z.

Caleb E.

Absolute Value - Example 1

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Ann Xavier GantertAlgebra 2 and Trigonometry

Grace H.

Anna Marie V.

Heather Z.

Caleb E.

Ann Xavier Gantert

Algebra 2 and Trigonometry