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In $9-26,$ solve each quadratic equation by completing the square. Express the answer in simplest radical form.$$\frac{1}{2} x^{2}+x-3=0$$

$x=\pm \sqrt{7}-1$

Algebra

Chapter 5

QUADRATIC FUNCTIONS AND COMPLEX NUMBERS

Section 1

Real Roots of a Quadratic Equation

Equations and Inequalities

Quadratic Functions

Complex Numbers

Polynomials

Campbell University

Oregon State University

McMaster University

Lectures

01:32

In mathematics, the absolu…

01:11

04:44

In $9-26,$ solve each quad…

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03:06

04:27

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02:34

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03:53

Okay, This question would like us to solve the following quadratic. And we're going to do this by completing the square, which is our way of transforming a quadratic that we can't really solve easy into one that we can solve by introducing these constants D and E, which are specially chosen to make this work out so d is be over to a and e is equal to C minus B squared over for a So now we just need to find our values of D and D. So D is equal. To be over to a or B is one and a is 1/2 so d is just one, and then e is equal to C minus. B squared over for a which is negative three minus one over four times 1/2 which is to So we get negative six halves minus another have and that's just equal to negative seven house. So after this is done, we can just plug into our Vertex form of 1/2 x minus or sorry, plus one quantity square minus seven halves. And now this equation is much easier to solve than what we started with originally. So let's just plug in. Be about 1/2 times X plus one quantity squared is equal to seven halves, or X plus one quantity squared is equal to seven or X plus one is equal to plus or minus the square root of seven. So now we have two branches for our solution. We have X Plus one equals Route seven and experts. One equals minus Route seven. So we get X equals. Weren't seven sorry negative one plus Route seven and X equals negative one minus Route seven. And these are our final answers.

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