Help Me Out With This Maths Question Be More Specific .

How do I solve this math question?

First let's try to develop a general theory for these kinds of questions. So lets consider a general equation of  Conic section which is [math]ax^2+2hxy+by^2+2fx+2gy+c = 0[/math]Rearrange this equation as [math]ax^2+2x(hy+f)+(by^2+2gy+c) = 0 [/math]Solving for the [math] x[/math] we have [math]x= \dfrac{-2(hy+f)\pm2\sqrt{(hy+f)^2-a(by^2+2gy+c)}}{2a} [/math]Now if we want that given equation should have linear factors then this also means that above expression should have linear relationship between [math]x[/math]  and [math]y[/math]. Now for this to happen the term under square-root should be a perfect square.[math]\implies (hy+f)^2-a(by^2+2gy+c)[/math] should be perfect square Simplifying this equation and we will get [math](h^2-ab)y^2+2y(gh-af)+(g^2-ac) = 0 [/math]Now this equation will be a perfect square if it has only one root i.e it's discriminant is zero.[math]\implies 4(gh-af)^2 - 4(g^2-ac)(h^2-ab) = 0[/math] Simplify this and we will get [math]\boxed{af^2+ch^2+bg^2-2fgh-abc = 0}[/math] Hence this condition should be true if you want to factorize any such equation into two linear factor. More neat and aesthetic form to express this condition is [math] \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c  \end{vmatrix} = 0  [/math]. This is also called discriminant of Conic section . (It decides whether the given equation represent an ellipse or parabola or circle or point or hyperbola or pair of straight line. Here it represents the pair of straight line).Now come to this particular equation. (I am changing [math]a[/math] to [math]P[/math] and [math]b[/math] to [math]Q[/math] to avoid confusion).So given equation is [math]3x^2 + 2Qxy + 2y^2 + 2Px - 4y + 1[/math]. If you compare this to standard form you will find that ,[math]a= 3 , h = Q , g = P , b = 2 , f = -2[/math] and [math]c = 1[/math] So discriminant of this equation is [math] \begin{vmatrix} 3 & P & Q \\ Q & 2 & -2 \\ P & -2 & 1 \end{vmatrix} = 0[/math]Solving and rearranging this we have ,[math]Q^2+4PQ + 2P^2 + 6 = 0 [/math]Clearly [math]q[/math] is a root of quadratic equation given by [math]x^2+4Px+2P^2+6[/math]

How do I solve this ACT math question?

I’m glad this is squares. Hexagons would be awful!This question brings together four math skill from an American high school course geometry & algebra:geometric knowledge of squares,the concept of “difference” as in subtraction,how to simplify radicals (aka square roots)arithmetic that’s allowed with simplified radicalsBegin with Skill 1. Given the area of a square, the side length is the square root of the area. Given a square of 50, the side length is √50. The smaller is √18. No need to compute the decimals with a calculator.Use Skill #3 how to simplify radicals: factor 50 to 2 * 5 * 5. Note that the 5 * 5 can be removed from the radical symbol, meaning √50 can be re-written as 5√2. The same can be done with √18. Factorized 18 is 2 * 3 * 3, and this, √18 = 3√2.Skill #2 informs us that x is the difference, or subtraction of 5√2 and 3√2.This is where most students might give up, unless they have Skill #4. If you know this related idea of 5n–3n = 2n, then you also know 5√2 - 3√2 = 2√2. Like terms work like that. Didn’t know that numbers with √2 behaved like that? They do!Notice this question was designed in such a way that there was NO need for a fancy scientific calculator. What the question is testing is seeing if you can bring multiple skills together- and two of the four skills (#1 and #2) are taught to kids in middle school (grades 6,7,8).Answer is G, 2√2.

What are some tips/tricks to solve math questions of the IIT JEE quickly?

Maths questions are easy but you can apply few tricks in order to solve the questions just by a simple method.If you get long multiplication questions like 855 * 142*311It seems to be a long calculation. But you have a trick to solve it . Just take a look at the unit digits. Yes, you got it right.5*2*1 =10 i.e unit digit is 0. If the options are in different unit digits than it is just easy for you.When you get the question to find a series than you simply need to apply numbers n upto 3. Than you would get the answer.The sum of the first n terms of the series1^2+2 . 2^2 + 3^2 +2 . 4^2 +5^2+2 . 6^2 + …………. is:n(n+1)^2⁄2 when n is even.When n is odd the sum is :A.3n(n+1)/2B.[n(n+1)/2]^2C.n(n+1)^2⁄4D.n^2*(n+1)/2Without even bothering to see what the question actually is let us put a value for n,ie; n=1 (note that this value should be odd)Now the required sum = 1^2 (the first term of the series only)= 1Now lets check in the optionsA.3n(n+1)/2 =31(1+1)/2=3 CANT BE THE ANSWERB.[n(n+1)/2]^2=[1(2)/2]^2= 1C.n(n+1)^2⁄4 =1D.n^2(n+1)/2=1Now we see that the options B,C,D all give us the correct answer.But only one of them is correct.Hence we need to put another value for n.Lets put n=3 and check.The required sum is 1+8+9=18A.3n(n+1)/2 Already eliminatedB.[n(n+1)/2]^2=[3(4)/2]^2= 36 therefore not the answerC.n(n+1)^2⁄4=36 therefore not the answerD.n^2*(n+1)/2=18Thus D is the correct answer.Make Algebra your weapon. Create the ability to picturize functions as graphs and learn to apply vertical and horizontal origin shifts carefully.Integral Calculus can be scoring if you use tricks and some basic varieties of integral functions. You can save a lot of time by using properties and applying them wisely.Work on these aspects and it would be easy for you to solve questions. Even if you need further help, than you can apply for online courses like Kaysons as they provide you all kinds of study materials and also give guidance to solve using tricks.