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Find the unknown.$$\frac{1}{8} x^{2}+3=0$$

$$\pm 2 \sqrt{6} i$$

Algebra

Chapter 0

Reviewing the Basics

Section 2

Solving Equations of the Form $a x^{2}-b=0$

Equations and Inequalities

Missouri State University

Oregon State University

Baylor University

University of Michigan - Ann Arbor

Lectures

00:58

Find the unknown.$$8 x…

01:07

Find the unknown.$$3 x…

00:50

Find the unknown.$$x^{…

00:56

02:36

Find the unknown.$$(6 …

01:44

Find the unknown.$$(x+…

01:43

Find the unknown.$$(x-…

02:17

Find the unknown.$$\fr…

01:30

01:23

00:33

Solve.$x^{2 / 3}-2 x^{…

02:06

00:40

Determine the unknown.…

03:08

Solve.$$3 x^{2}-8 …

we're solving for the unknown, which is one uh what? Sorry. The unknown as ex. Excuse me. Um, so we're solving for X and what I would do. It's just try to get X by itself. Um, before you get X by itself, try to get X squared by itself. So you can start by subtracting three over, which is pretty straightforward. Zero minus three is negative three. And for right now, I'm gonna leave. This is 1/8 um, but instead of dividing both sides by 1/8 and done, a lot of students preferred, You know, it's a multiplication problems that they wanted to buy. But dividing by a fraction is the same as multiplying by the reciprocal. So I multiply both sides by 8/1. And I am well aware that eight of the one is just eight. But you can actually see the eights, and the ones cancel out. So you're left with X squared on the left side is equal to negative 24. Now, your last step is to square root. Now for some books, we don't want to deal with imaginary numbers. So some books might say that there is no real solution here, and if you're confused by that quick explanation is, there's no way of taking a real number of positive and negative squaring it and getting a negative because even the negative squared is positive. But we do have this tool, Uh, and it's the what is an imaginary number. So we used the letter I to represent imaginary number and you can break down 24 leaves red as four times six and then four breaks down as two and two. Now you could break down six as two and three, but there's no other pairs were looking for pairs here, so our final answer is we can bring the pair of twos in front Route six. And the reason why that trick works is what I'm technically doing is I'm rewriting, um, Route 24. 1st of all, let's get that high in front as a square to four times the square root of six, and we know the square to four is, too. So that's why that trick works. And this is our final answer. Plus or minus two I Route six

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