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Find the unknown.$$6 w^{2}+96=0$$

$$\pm 2 \sqrt{3}$$

Algebra

Chapter 0

Reviewing the Basics

Section 2

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

Equations and Inequalities

Missouri State University

Campbell University

Harvey Mudd College

Lectures

01:05

Find the unknown.$$6 w…

00:46

Find the unknown.$$w^{…

02:29

Find the unknown.$$\fr…

02:58

Solve.$w^{4}-6 w^{2}+2…

01:43

Find the unknown.$$(x-…

01:02

Solve each equation.$$…

02:48

Solve the equation.$$w…

07:01

Solve using the quadratic …

05:17

$$w^{\prime \prime}+4 w^{\…

00:17

Simplify.$$6-w^{0}$$

01:41

Solve the quadratic equati…

02:06

Solve each equation and ch…

00:23

all right. So as we're looking at this problem and they're just asking you to solve for W, that's the unknown in this problem. What I would do is start by getting W by W squared by itself. So you can do that by first subtracting 96 over. That's pretty easy, because zero minus 96 would be negative. 96. I'll leave this six still here for a second, and then to get six over, you gotta divide both sides by six. So just using a calculator and the negative 90 65 by six is negative. 16. Uh, Now, the quick answer at this point is that you cannot. There's not a real solution because there's no way of getting a a number when you square it to give you a negative number. So there's no real solution in this problem, so I'm gonna circle that. But if you take it a step further and you square root both sides, um, you can pull out an eye because that is equal to the imaginary number is the square root of negative one. The square is 16, is four and then don't forget, you can have a plus or a minus positive negative. Four I. So this is the context. They want this answer. Go ahead and circle that. But your teacher might also. So this is your complex solution, but your teacher might want you to recognize that there's no real solution. Um, I would do whatever your professor months, but anyway, plus or minus four, I should be good.

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