Видео

Hi - Honestly I have no idea how to send them to Binance. :( Thank you for your support!

Thank you for your support!

Thank you for your support!

Thank you for your support!

Thank you for your support!

Hi - Yes, we can apply it for quadratic equations, but we need to understand the possible outcomes. Scenario# 1: Two quadratic equations having one common solution (root). ***************************************************************************** a) When the leading coefficients are equal, equating the quadratic functions will give us a linear equation that will essentially give us the common solution. Let's take an example:- x^2 + x - 2 = 0.........(i) x^2 - 3x + 2 = 0.......(ii) By equating the quadratic functions, we get: x^2 + x - 2 = x^2 - 3x + 2 => 4x = 4 => x = 1 This is the common solution (root) for the original equations. b) When the leading coefficients are different, equating the quadratic functions will give us another quadratic equation that will also have the same common root. Let's take an example:- x^2 - 3x + 2 = 0.........(i) 2x^2 + 5x - 7 = 0......(ii) By equating the quadratic functions, we get: x^2 - 3x + 2 = 2x^2 + 5x - 7 => x^2 + 8x - 9 = 0........(iii) This new equation (equation (iii)) also has the same common root: x = 1. Now, we can again take equation (i) and equation (iii) to create another new equation which will also have the same common root. Similarly, we can take equation (ii) and equation (iii) to create another new equation which will also have the same common root. This way, we can create infinite number of equations that will have the same common root. Scenario# 2: Two quadratic equations having two common solutions (roots). ******************************************************************************* a) When the leading coefficients are equal, the equations will be identical. Equating the quadratic functions will lead to an equation like this: 0 = 0 b) When the leading coefficients are different, one quadratic function will be a transformation of the other one. Eventually, we will be able convert them into identical equations. For example:- x^2 - 3x + 2 = 0.........(i) -2x^2 + 6x - 4 = 0.....(ii) By equating the quadratic functions, we get: x^2 - 3x + 2 = -2x^2 + 6x - 4 => 3x^2 - 9x + 6 = 0 (by bringing everything to the LHS) => 3(x^2 - 3x + 2) = 0 => x^2 - 3x + 2 = 0......(iii) This new equation (iii) is identical to the original equation. We can even convert original equation (ii) to equation (i) :- -2x^2 + 6x - 4 = 0 => -2(x^2 - 3x + 2) = 0 => x^2 - 3x + 2 = 0........(iv) This new equation (iv) is also identical to the original equation. So, when both roots are common, the original equations are essentially identical. Hope it is clear! Thank you for your support!

Hi - Every linear equation in 'x' & 'y' represents a straight line. You can equate the LHS of the two original equations. This will give you a new equation which will represent a straight line that goes through the point of intersection of the two original straight lines. Let's take a example here:- x + 2y - 7 = 0.........(i) 2x - 5y + 4 = 0.........(ii) These 2 equations represent a pair of straight lines that intersect at the point (3,2). Now, let's equate the LHS of the equations. We will get an equation like this:- x + 2y - 7 = 2x - 5y + 4 If we simplify it, the equation will look like this:- x - 7y + 11 = 0.........(iii) This new equation also represents a straight line which goes through the point of intersection of the original 2 lines and the common intersection point is (3,2). The new equation is also satisfied by x = 3 and y = 2. From (i) and (iii) you can create another equation. From (ii) and (iii) also you can create yet another equation. In fact, you can create infinite number of equations this way. All of those straight lines will go through the same common intersection point. Hope it is clear!

Hi - You are absolutely correct! I will make a separate video on this particular topic when I cover Complex Numbers. Thank you for your support!

@@finemath You are welcome.

Thank you for your support!

Thank you for your support!

Hi - You are absolutely correct in your approach! In the final step, while bringing out (ax1 + by1 + c)^2 outside of the square root, you need to use the absolute value sign. I mean, square root of (ax1 + by1 + c)^2 is actually equal to | (ax1 + by1 + c) | . Great job! Thank you for your support!

Hi - You are absolutely correct! I have called it the 2nd Method and here is the link for that video: ruclips.net/video/UIz87RMXQOQ/видео.html Thank you for your support!

@finemath Thanks a lot.

Thank you for your support!

Thank you for your support!

When you solve two simultaneous linear equations, the solution represents the location (co-ordinates) of the point of intersection of the two straight lines represented by the two equations. You can equate the LHS of the original equations if their RHS values are same (as in your example). But, when you do that, you are going to get a new linear equation which will look like this : ax + by = px + qy, => (a - p)x + (b - q)y = 0, and this new equation would represent a straight line which passes through the Origin & the point of intersection of the two original straight lines. That means, you now have a 3rd simultaneous equation, and the three straight lines have a common point of intersection. Solution of any two of those three equations will give you the location of the point of intersection of those three lines. If equating the LHS (when the RHS is same, as in your example) helps you find the solution easily, feel free to do it. Hope I answered your question. Thank you for your support!

​​@@finemathI am asking because mobile keyboards do not give a proper "symbol of math" and some sign looks same . sir In your answer [new linear equation which will look like this : "ax + by = px + qy, => (a - p)x + (b - q)y = 0" ] you are using the "(=>)" sign in this equation. This sign (=>)shows "implies" OR "greater than equal to" here Please please reply .

​@@finemathsir In your answer [new linear equation which will look like this : "ax + by = px + qy, => (a - p)x + (b - q)y = 0" ] you are using the "(=>)" sign in this equation. "What does it mean here . I hope you can help

Hi - The "=>" sign means 'it implies that" or "hence".

Hi - The "=>" sign is not a "greater than or equal to" sign. This sign is used to indicate: 'it implies that" or "hence".

Hi - Thank you for your feedback! I have now added a general (formal) proof in the description of this video. Please review and let me know if it helps or if you need any clarifications. Thank you for your support!!!

Thank you for your support!

Thank you for your support!

Hi - Here's the link for the video on Cross-Multiplication method: ruclips.net/video/p1nu41ZAU5o/видео.html Let me know if that's what you are looking for, or something else. Thank you for your support!

Thank you for your support!

You are absolutely correct!!! Thank you for your support!

Thank you for your support!!!

Thank you for your support!!!

Thank you for your support!!!

Excellent! Thank you for your support!!!

Thank you for your support!

Thank you for your support!!!

Thank you for your support!

Thank you for your support!

Thank you for your support!

Thank you for your support!!!

Thank you for your support!

Thank you for your support!

Thank you for your support!

Thank you for your support!

Thank you for your support!

Hi - If you are talking about 140° as in interior angle of a regular ploygon, then we can try this formula to find the type of the regular polygon / no. of sides. Each Interior angle of a regular polygon = ((n-2)*180°)/n ⇒ 140° = ((n - 2)*180°)/n ⇒ 140°.n = (n - 2)*180° = n.180° - 360° ⇒ n(180° - 140°) = 360° ⇒ n.40° = 360° ⇒ n = 360°/40° = 9 So, the no. of sides of the regular polygon is equal to 9, which means it is a Nonagon. Thank you for your support!

Hi - Thank you for your suggestion! After completing Conic Sections, I am planning to start working on topics from Algebra. After completing Algebra, I will start working on Vector and 3-D Geometry. After that, I will move on to Precalculus & Calculus. Along with Precalculus, I will cover Inverse Trigonometric Functions & Trigonometric Inequalities. Thank you for your support!

Thank you for your support!!! Please feel free to share my channel with your friends!

Thank you for your support!

Thank you for your support!

Hi - Thank you for the kind words! If you carefully look at △CEB, it is actually a right triangle because ∠CEB is a right angle since CE is perpendicular to AB. In a right triangle, the Cosine ratio of an acute angle = adjacent side / hypotenuse. In the right triangle △CEB, the adjacent side for ∠CBE (which is equal to 60°) is EB and the hypotenuse is BC. That's why, in △CEB, Cos 60° = EB/BC. Hope it is clear now. Appreciate your blessings & support !!!

​@@finemath You're welcome, anytime for the kind words! Wow, I'm stunned at how well you expounded it in the video and how I somehow didn't understand it the first time; when I was in the middle of reading your compendious explanation, I scrolled back to the video, it all clicked, and I solved the problem in about 25-30 seconds. It's crystal-clear now, for you made it a seemingly difficult problem become relatively easy! You're very welcome; your pure-hearted help is admirable and exemplary, so please firmly grasp it forevermore!

Thank you for your support!

Thank you for your support!

Hi - Thank you for your support & feedback! I am planning to complete the theory portion first. After that, I will do the PYQ's.