Math Problems Quite Confused Mum Here Help

I am very weak in maths. How can I develop it?

I'm very weak in Maths, or I don't know whether it's something that I, myself developed over the years. From my 7th grade, I started loosing interest in Maths since my teacher wasn't that good and she only appreciated students who did well and insulted us very badly. From that year, I just gave up my hopes on Maths and considered it as a shitty subject and pretended like I didn't care when I was really really very sad on getting very low marks in Maths when I was an excellent student in the class and scored more than 91% in other subjects, Maths always bought me down when it came to ranks. I was so depressed, I started hating Maths, and it became a nightmare for me. The day before the exam, all I remember doing was crying, crying and crying. I couldn't bear the tension it gave me. It still does. I suffer from anxiety even though I'm young; and no one is allowed to put so much pressure on me because it's trouble. Whatsoever, I still have chills and I failed in this year examination and writing a retest on Monday, and this time, I just googled whether there's anything I could do to score better in Maths, and that's when I read this interesting article and found out that I'm not alone. Because this is exactly like my story. I'm weak, I always get the just-pass marks, my brother is a top-scorer who is hopefully cracking a AIIMS soon and has 95% marks in his 12th grade and all, and I left a group tuition cause I was insulted and then joined a tuition where I used to be the only student, and I'm continuing that, but the thing I failed to keep is being regular. I'm never regular to the tuition because I hate maths firstly and I'm a lazy girl who spends her weekend by watching new movies, with drinks and chips and popcorn and stuff and going out every weekend for shopping with my family or friends, or going to stay at cousin's or mom's or spending quality time with my best friends or my best-friends-cum-family my cousins..Okay! I've told too much. I hate maths. No, I hated it. Not anymore. I'm going to pretend like I love it and work hard. I've got 5 days from now, that's like huge time to cover 6 lessons and then practice extra questions and sums. Well then! Let's see how I'll score and whether the hardwork will pay off! Fingers crossed! Maths!! HERE I COME BROTHER.

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]