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Find the unknown.$$(6 x-4)^{2}+48=0$$

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

Algebra

Chapter 0

Reviewing the Basics

Section 2

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

Equations and Inequalities

McMaster University

University of Michigan - Ann Arbor

Lectures

01:43

Find the unknown.$$(x-…

00:56

Find the unknown.$$x^{…

00:58

Find the unknown.$$8 x…

01:07

Find the unknown.$$3 x…

02:17

Find the unknown.$$\fr…

00:33

01:05

Find the unknown.$$6 w…

01:40

00:51

$$\text { Solve } 4(x-3)^{…

00:44

Solve the equation.$$4…

00:47

00:32

00:50

01:12

Solve.$$x^{4}-x^{3…

So we're solving for X in this problem. Six X minus four squared plus 48 is equal to zero. So what we need to do is get this squared term all by itself. And we do that by subtracting 48 over. So zero minus 48 is negative. 48. So this left side is all by itself. And, uh, we need to undo the square at this point by square rooting both sides. So what's important to recognize is the square root in the square root cancel. So we're left with six X minus four on the left side and on the right side. Uh, when you square root and negative or first of all square root and you get plus or minus answer, you have to pull an eye in front you square root. Now the way I would do this is I will break this down. I'll just keep it as simple as 2 24. Breakdown 24 to 12. Breakdown 12 as two and six and six is two and three and any pair of numbers I can bring in front. I'm looking at two times. There's another pair of tooth two times two. I route the non pair over here. Um, now, it would have been faster if I rewrote. This is I square root of 16 times. The square root of three is the square root of 16 is four. So what are we looking at? We're looking at six X minus four is equal to plus or minus four. I route three. Well, now we can solve for X by doing two steps. The first step is adding forward to the right side. And that's a real number. That's a whole number. So we cannot combine it with the imaginary number of here and to solve the multiplication problem you divide. Now the only thing is when all of the whole numbers are divisible by the same number 44 and six, they're all divisible by two. The expectation is that you do divide them all by two. So four divided by two is two four. Divided by two is two. We don't mess with the radical because that's not a whole number. Irrational. Uh, and six divided by two is three. So this is your best answer for this problem? Completely simplified

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