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Master Square Root Estimation: Quick & Accurate Techniques

8.EE.A.1

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Using exponents to describe numbers

Expanded powers

Exponent rules

Single Power Expressions

Dividing powers

Multiplying powers

Order of operations with exponents

Product rule of exponents

Quotient rule of exponents

Power of a product rule

Power of a quotient rule

Power of a power rule

Simplify the power

8.EE.A.2

Squares and square roots

Square Root of a Perfect Square

Square and square roots

Cubic and cube roots

Cube root of a perfect cube

Converting radicals to mixed radicals

Converting radicals to entire radicals

8.EE.A.3

Scientific notation

8.EE.B.5

Parallel and perpendicular lines in linear functions

Understanding graphs of linear relationships

Linear relations

Understanding tables of values of linear relationships

Linear relation

Applications of linear relationships

Identifying proportional relationships

8.EE.B.6

Slope equation: \(m = \frac{y_2-y_1}{x_2- x_1}\)

Slope equation: <katex>m = \frac{y_2-y_1}{x_2- x_1}</katex>

Slope intercept form:  y = mx + b

Slope intercept form: y = mx + b

8.EE.C.7

Introduction to linear equations

Model and solve one-step linear equations: ax = b, x/a = b

One-step Linear Equations

One Step Linear Equations — Word Problem

One-step equations

One-step Equations LHS

One-step Equations RHS

Model And Solve One Step Linear Equations Divide – Word Problem

Model And Solve One Step Linear Equations Multiply – Word Problem

Solving two-step linear equations using addition and subtraction: ax + b = c

Two-step Linear Equations

Two-step Linear Equations LHS

Solve Linear Equations Using Addition and Subtraction – Word Problem

Two-step Linear Equations — Word Problem

Solving two-step linear equations using multiplication and division: x/a + b = c

Two-step Linear Equations RHS

Solving two-step linear equations using distributive property: a(x + b) = c

Solve Linear Equations Using Distributive Property – Word Problem

Solving literal equations

8.EE.C.8

Determining number of solutions to linear equations

Solving systems of linear equations by graphing

Solving systems of linear equations by elimination

Solving systems of linear equations by substitution

8.F.A.1

Relationship between two variables

Identifying functions

8.F.A.2

Function notation

8.F.B.4

Applications of linear relations

8.F.B.5

Graphing linear functions using table of values

8.G.A.1

Line symmetry

Tessellations using translations and reflections

tessellations using translations and reflections

8.G.A.2

Rotational symmetry and transformations

8.G.A.3

Enlargements and reductions with scale factors

Tessellations using rotations

8.G.A.4

Similar polygons

8.G.A.5

Similar triangles

8.G.B.6

Pythagorean theorem

Pythagorean Theorem

8.G.B.7

Using the pythagorean relationship

8.G.B.8

Distance formula: \(d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

Distance formula: <katex>d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}</katex>

Distance formula

Applications of pythagorean theorem

8.G.C.9

Volume of cylinders

Word problems relating volume of prisms and cylinders

Surface area and volume of cones

Surface area and volume of spheres

8.NS.A.1

Understanding the number systems

Rational vs. Irrational numbers

8.NS.A.2

Determine square roots of rational numbers

Square roots of rational numbers

Estimating square roots

Converting repeating decimals to fractions

8.SP.A.1

Reading linear relation graphs

Box-and-whisker plots and scatter plots

8.SP.A.2

Solving linear equations by graphing

8th Grade Math

math

Intro

- [Instructor] Hi, welcome to this question right here. So we're trying to estimate what is root 66 manually, now without the calculator. So the work is actually pretty simple. You draw yourself a number line and you make yourself two lines right here to represent two numbers. Now these two numbers here is going to be enclosing the 66 but they have to be perfect square numbers. For example, 64 is... that's the first number that come to me because 66 is greater than 64, right? Sixty-four turns out to be and eight square which is a perfect square number. So I know the root 66 is definitely greater than eight but eight point something, something which I don't know yet. Then this number over here is going to be... after eight there's a nine, nine times nine is 81. So this is 81 right here. So I know this root 66 is definitely between 8 and the 9. But where is it? 
Well, we can actually calculate it approximately, estimation. Take the 81 right here and subtract to 64. There's 17 units across from 64 to 81. 17, so I'm not going to do all of them though. I'm just going to cut it in half. So this will be 8.5, and if there's 17 units. 17 divided by 2 I get 8.5 units. So 64 + 8.5, I got myself-- This is actually my 72.5, on the number line, but in terms of the square root of this number. Let me put square numbers on these guys right here. It will be 8.5, just perfectly halfway. Now, the 66 is actually between these two guys right here, but are they closer to the half or they're closer to the 8? 
Well, if I just take my 72.5 over here and subtract the 64, I have 8.5 units. So this one is actually not that bad. I can actually do 8.5, then I go-- That's halfway, that's 1, 2, 3, 4 5, 6, 7, 8 and a half maybe. This is approximation, so don't be too serious about these guys. Approximately 8.5 units. Okay, then I think-- Every unit is a one now, so that's 66 right? This is 64, 65, 66 is right here. This is my 66, okay? So that's root 64, root 65, root 66. So as you can see, it's very, very close to my 8. See, this is 8 right here. This is 8.5, so if you just quickly do a quick approximation. It will be 8.1, 8.2, 8.3 and 8.4. If I just quickly draw a lines approximately because I don't, I didn't really do have a good scale on these guys. If you have like a grid paper, probably much easier. This is 8 right? 
Then this is like 8.1, 8.2, 8.3, 8.4, 8.5. So it looks like I'm just a little bit on the 8.1 something. I'm going to do an approximation, 8.1. That's pretty much it, 8.10. That's what I'm going to put down. Approximately estimated, it's going to be my 8.10. Now, if I punch it in a calculator, radical of 66 then I got myself-- I'm pretty close actually. It's going to be 8.12, but once again I don't have a good scales right here. If you draw them perfectly with the equal distance, if you draw the units perfectly, you will get a better approximation than this. But that's the work, and once again find your perfect score numbers that will enclose that number 66, and then just draw, figure out how many units between them, figure out how many units between them, then just slowly figure out where your location is and just do the approximation. Thanks for watching.

To estimate square roots which are not perfect squares without using a calculator, we need to know the perfect square numbers well. We will first put the number inside the square root sign in the middle of a number line, and then find the two closest perfect square numbers on its left and right hand side to make the best estimation.

How do you estimate the value of square roots?

Divisibility rules

Introduction to Exponents