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Find the unknown.$$(y-5)^{2}+12=0$$

$$5 \pm 2 \sqrt{3} i$$

Algebra

Chapter 0

Reviewing the Basics

Section 2

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

Equations and Inequalities

Missouri State University

Campbell University

Oregon State University

Idaho State University

Lectures

01:29

Find the unknown.$$5 y…

01:59

Find the unknown.$$(2 …

00:24

Find the unknown.$$y^{…

01:05

02:26

Find the unknown.$$\fr…

00:36

Find the unknown.$$2 y…

01:35

Solve the equation. $$…

Solve. $(3 y+5)^{2}=0…

01:00

Solve.$$y 2-y+5=0<…

our task is to solve for Y in this problem. So it wasn't that y minus five squared plus 12 is equal to zero. Uh, so when you're solving a problem like this, you want to get the squared all by itself. In other words, I need to get rid of this plus 12, which I can do that by subtracting 12 to the right side. Um, zero minus 12 is negative 12. And this y minus five squared is all by itself, like I want now. A quick explanation is that it's impossible for a number when I square it to give me a negative because even negative squares are positive. Negative, positive. So we're going to get a non real answer. Unreal solution. But what we can also do is we can we can square root and we can use this complex number plus or minus, um, square root. Now, the way I'm going to show this is I'm going to factor tree. 12 down is four and three and two and two, so you can bring a pair of twos in front. I'm well aware that some teachers would prefer seeing well, you have the I, uh and rewriting as the square to four times the square of the three because the square root of four is, too. But that's another teaching strategy for you and on the left. I just have that y minus five. So to solve for, For why is to add five to the right side and because that's a real number, the whole number we cannot combine it with this imaginary, this complex piece. So I just leave our answer like this Mhm.

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