Competitive Math Made Easy

An equation, in which the highest power of the variable is $2$ is called a quadratic equation. The standard form of a quadratic equation is $ax^2+bx+c=0$, where $a, b,$ and $c$ are constants and $a \neq -$ Solution of Quadratic Equations: There are two main methods to find the solutions, or roots, of a quadratic equation. Factorization $\rightarrow$ Let $ax^2+bx+c = a(x - \alpha)(x - \beta) = 0$. Here, the roots of the equation are $x = \alpha$ and $x = \beta$. Hencec, factorizing the equation and equating each factor to zero is one method to find the roots. The second method is known as the Quadratic Equation, which is:  $$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ where the equation of the quadratic is $ax^2+bx+c = 0$.  Nature of the Roots:  In a quadratic equation in standard form, the term $b^2-4ac$ is known as the discriminant of the equation, which plays an important role in finding the nature of the roots. It is often denoted by $D$. If $a, b, c \in \ma...

In $\triangle ABC$, let the name the incenter as point $I$ and the feet of radii as $D, E,$ and $F$ as shown below. Since $I$ is the incenter, $ID = IE = IF = r$ Now, using the $A = \frac{bh}{2}$ formula on $\triangle ICB, \triangle IAC, \text{and } \triangle IBA$, we get   $[ICB] = \frac{1}{2}ID \cdot BC = \frac{ra}{2}$ $[IAC] = \frac{1}{2}IE \cdot AC = \frac{rb}{2}$ $[IBA] =  \frac{1}{2}IE \cdot AB = \frac{rc}{2}$ Adding all three of our equations, we have  $$\frac{ra}{2} + \frac{rb}{2} + \frac{rc}{2} = [ABC]$$ Letting $s = \text{semiperimeter} = \frac{a+b+c}{2}$, we get $$[ABC] = r \cdot s$$

Stewart's Theorem states:  $$t^2=\frac{b^2u+c^2v}{a}-uv$$ The theorem can also be written as: $$b^2u+c^2v=(a)(uv+t^2)$$ Proof: In $\triangle AED$, by Pythagorean theorem, we have $x^2+h^2=t^2$                                                                                                                                    (1) Similarly, in $\triangle AEC$ $h^2+(v-x)^2=b^2 \Rightarrow h^2u+uv^2-2uvx+ux^2=b^2u$                                               (2) In $\triangle AEB$, we use Pythagorean theorem again and get ...

Factorization is the process of writing a number or expression as a product of several factors, usually smaller or simpler. For example $8$ can be factorized as $2 \cdot 4$, and $(x-3)(x+3)$ is the factorization of the polynomial $x^2-9$. Here are some common factorizations worth memorizing: $$(a-b)(a+b)=a^2-b^2$$ $$(a \pm b)^2 = a^2 \pm 2ab + b^2$$ $$(a \pm b)(a^2 \mp ab + b^2) = a^3 \pm b^3$$ $\text{Proof }$ $$(a+b)(a^2 - ab + b^2) = a^3 - a^2b + ab^2 + a^2b - ab^2 +b^3 = a^3 + b^3$$ $$\text{Use ($-b$) to replace $b$ in the above, we obtain,}$$ $$(a-b)(a^2 + ab + b^2) = a^3 - b^3$$ $$(a \pm b)^3 = a^3 \pm 3a^2b + 3ab^2 \pm b^3$$ $\text{Proof }$ $$(a+b)^3 = (a + b) \cdot (a + b)^2 = (a+b)(a^2+2ab+b^2)$$ $$(a+b)^3 = a^3 + 2a^2b+ab^2+a^2b+2ab^2+b^3$$ $$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$$ $$\text{Use ($-b$) to replace $b$ in the above, we obtain,}$$ $$(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$$ $$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$$ $\te...

Probability is a quantity that expresses the chance, or likelihood, of an event. It is most helpful to think of probability as a fraction. The literal definition of probability is the chance of occurrence of an event. For example, if a person is standing at the intersection of two roads which direct towards North, South, East, and West. Thus, he has a total of $4$ choices (four different directions) to proceed. Now, if he wished to go towards a particular direction, then the probability of completing his wish is $\frac{1}{4}$ since he can only choose one out of the four directions. Consider another example: A person has two different cars, a Toyota and Honda, which he uses randomly. It can then be said that the probability of using the Toyota is $\frac{1}{2}$ because out of his total of $2$ cars, he can randomly pick $1$ of them. Hence, from the above examples, we can conclude that the probability of an event occurring is $$ = \frac{\text{Number of desired or successful outcomes...

Let $A, B, C$ be three letters , then we can combine any two of them in the following ways: $$ AB, BC, AC$$ Similarly, if $A, B, C, D$ are four letters, then we can combine any two of them in the following manner: $$ AB, AC, AD, BC, BD, CD$$ Similarly, we can combine any $3$ of $A, B, C, D$ as : $$ ABC, ABD, ACD, BCD$$ We can generalize by saying that the number of all combinations of $n$ distinct things taken $r$ at a time $(r \le n)$ is ${n}\choose{r}$  $= \frac{n!}{(n-r)!r!}$ Note In combinations, the order of the letters (or things) is not considered. Here, AB and BA are the same, so they are only counted once, unlike permutations. The term "combination" is generally used for selection of things and "permutations" are used for rearrangements. Combinations with Restrictions Number of combinations of $n$ things taken $r$ at a time in which $x$ particular things always occur is $${n-x}\choose{r-x}$$ Number of combinations of $n$ things t...

In circular permutations, things are to be arranged in the form of a ring or a circle, e.g. arrangements of people around a circular table. In circular permutation there are no end points, i.e. there are no beginning or ending positions. So, the number of circular permutations of n objects are $$\frac{n!}{n}=(n-1)!$$ Thus in a circular permutation, one thing is kept fixed and the remaining  $(n-1)$ things are arranged in $(n-1)!$ ways. If the clockwise and counter clockwise orders are not distinguishable, then the number of ways = $\frac{1}{2}(n-1)!$ Let us look at a few examples.  Ex:   In how many ways can $6$ boys be seated at a circular table? Sol:   We keep one boy in a fixed position and to find the number of permutations to arrange the remaining 5 boys, we simply take $5!$, which gets us to our answer $\boxed {120} $.  Ex:  In how many ways can $6$ boys be arranged at a round table so that $2$ particular boys can be seat...