 Hello friends welcome again to another session on number system and in this session. We are going to discuss how to find out root of any positive number positive root of any positive number on the number line okay, so let us understand how to do that and so Basically, let us say I have to find out a root of 7.4 Okay, so this is what I am Intending to do so let us say I am trying to find out root of x root of x where x is equal to 7.4. This is my Objective how to do that? These are the steps look carefully and I'll also explain what is The rationale or why this particular thing would work So this works on the identity, which is this which is x plus 1 by 2 whole square Minus x Minus 1 by 2 whole square if you do this. This is nothing but equal to x Okay, so if you somehow find out Root of both then you will be able to find out root of x So that means if I put a square root over it that means this is equal to root of x So this is what the underlying philosophy of finding root of x on the number line Let us see how to do it the steps are like these first of all We have to draw a line segment of length 7.4 that is whatever you want to Find the square root of for example in this case it is x so I have to draw a line segment of the same length So this length is 7.4. Isn't it? So you can see a line segment of length 7.4 has been Drawn now again. You have to extend this line segment by one unit. So let us say now I am drawing BC which is of length one unit So total length AC is nothing but 7.4 plus one now next step is to find a perpendicular bisector of AC So let us draw the perpendicular bisector of AC. So a See so now I have drawn perpendicular bisector of AC and let the point of intersection of the perpendicular bisector and AC is d D, okay This is d now the next step is to take d as the center and Radius as ad or dc. You have to draw a semicircle. So I am now drawing semicircle between a and c with d as the center Okay, so this is a semicircle with d as the center next is you take a line perpendicular to this AC and Passing through B. So I am drawing Yeah, so if you see this line now the draw line drawn is perpendicular to AC and passing through B and let us now Find a point which is the intersection of the perpendicular and the semicircle. So that point is now e This is e next is what you need to do is you need to do is take d as the center and E as the radius and before that let us just join d and e for our convenience So I am not drawing I'm joining d and e for our convenience because I Will be proving a little while later how this particular thing works So now taking B as center and be as radius again taking Be as center and be as radius So if you see I drew a circle with B as the center and e is that e is that be as the radius and now I am trying to find out this point of intersection of point of intersection of The circle and the x-axis and that point is f now. Let us do one thing Let us now measure what bf is so now if you see the length bf is given as 2.72 and if you have a calculator with you or you can do a simple calculation actually bf 2.72 is approximately Root of 7.4 which was our x isn't it so we wanted to find out root of 7.4 Which is now given as bf. So now you can treat B as the origin and Bf point f on the number line represents root of 7.4 Okay, so once again, what did we do we first of all draw we draw we drew a line AB With this which is a 7.4 unit long then we extend B to BC to see which is One unit away that is BC is one unit then we perpendicularly bisect AC to find point D Then on D as the radius and AD as sorry D as the center and AD as the radius We had drawn a semicircle a and This top point of the semicircle is C So AC if you see this one circle which is mentioned as small C So this is a semicircle which we drew then what did we do we took point B and drew up a perpendicular be on Passing through B which intersects the semicircle at point E Then taking B as the radius, sorry B as the center and be as the radius we drew a circle intersecting X-axis at F. So you need not draw a full circle You can simply draw an arc EF such that F is on the X-axis here F is in the X-axis Now the segment Bf represents the root of X and the proof for this is like Just described over here. So if you see let me write it down and explain it to you so if you see Points DB if you see what is DB or DE? DE is given as DE is nothing but the radius of the semicircle, isn't it? Which is nothing but half of AC 1 by 2 AC Isn't it 1 by 2 AC will be equal to nothing but AC is what X plus 1 So X was my AB the length AB was X plus 1 Divide by 2 is DE. So we got DE as X plus 1 by 2 now. What is Let us say DB now DB will be equal to DC DC which is equal to the radius again, which is nothing but X plus 1 by 2 and minus so DB is DC minus BC BC So this is X plus 1 by 2 minus than 1 because BC was one unit and if you do this calculation You'll get it is X minus 1 upon 2 Is it it so hence? hence BE BE Square it will be equal to equal to DE squared minus DB squared, isn't it? DE squared minus DB squared by what by Pythagoras theorem. So hence my dear friends BE will be equal to under root DE squared minus Dd squared Isn't it and if you look closely This particular relationship is nothing but what we have explained here. So D BE is nothing but X and DE is nothing but X plus 1 by 2 whole square and then DB is X minus 1 whole square and that is the that is the That is how we found out that this length BE represents root of X now since BE represents root of X, but I had to show it on the number line So what do I do I draw a circle or an arc taking B as the radius and BE as the sorry BE as the center and BE as the radius and that particular circle intersected the x-axis at F Okay, you don't need to draw a full circle even an arc will do but the point being that BE is equal to BF isn't it why because BE is the radius of the circle and BF is also the radius of the circle So hence BF in a way represents BE which is nothing but root X which was found out by Pythagoras theorem So this is how guys you can find out Any number square root though it was given in this example as 7.4 You could have done 9.7 9.3 8.11 11 whatever and you have to do exactly the same Right quick repetition draw a line AB of the length of which square root You want to find extend it by one unit to see draw the perpendicular bisector of AC find the point D taking D as the center and radius as AD make a semicircle and then Take the point B from point B draw a perpendicular intersecting the semicircle at E Now be as the center and be as the radius draw an arc intersecting The number line at F BF represents root of X. That's how the process is. Hope you understood the process. Thank you