## Tag Archives: Basic Algebra

### Basic Algebra for IITJEE Main and RMO

More basic algebra for you guys who are thirsting for more…The following is a nice problem indicating some basic concepts or tricks in problems involving logarithms/powers.

Solve for x: $4^{x}-3^{x-(1/2)}=3^{x+(1/2)}-2^{2x-1}$ (IITJEE 1978)

Solution:

Writing $4^{x}$ as $2^{2x}$ and bringing powers of the same number on the same side we get,

$2^{2x}+2^{2x-1}=3^{x+(1/2)}+3^{x-(1/2)}$

The first term on the LHS can be written as $2^{2x-1} \times 2$, and hence, a common factor of

$2^{2x-1}$ comes out from the terms on the LHS. As for RHS, we can rewrite the first term as

$3^{x-(1/2)} \times 3$ and then the factor $3^{x-(1/2)}$ comes out as common. So, we get

$2^{2x-1}(2+1)=3^{x-(1/2)}(3+1)$, that is, $3 \times 2^{2x-1}=4 \times 3^{x-(1/2)}$

Bringing all powers of 2 to  the left and all powers of 3 to the right, we get

$2^{2x-3}=3^{x-(3/2)}$

By inspection, $x=3/2$ is a solution. But, how do we arrive at it systematically? Also, how do we know that there is no other solution? It is tempting to try to do this by saying that a power of 2 can equal a power of 3 only when are both are equal to 1. (Such a reasoning is indeed useful in solving equations in Number Theory where we mostly deal with positive integers and their factorization into integers). But, here it is inapplicable because we do not  know that the exponents are integers. Instead, let us express both the sides as the power of the same number. One way to do this  is to write 3 as $2^{\log_{2}3}$ in the RHS. Then, we can get

$2^{2x-3}=2^{(\log_{2}3)(x-(3/2)}$

As the bases are the same, the equality of powers implies that of the exponents. So, we have

$2x-3=(\log_{2}3)(x-(3/2))$

This can be solved easily to give $x=\frac{3-(3/2)(\log_{2}3)}{2-\log_{2}3}$

which simply equals $3/2$.

Hence, $x=3/2$ is the only solution.

### Basic Algebra for IITJEE Main

Main Problem: For an integer $n \geq 5$, let $S_{n}$ denote the sum of the products of the integers from 1 to n taken three at a time. Then, what is the value of $S_{10}$?

First Hint: Consider the terms in the expansion of $(1+2+3+ \ldots +10)^{3}$.

Second Hint: Classify these terms according to the number of times they occur.

Solution: There are totally 1000 terms in the expansion of

$(1+2+3+ \ldots +10)^{3}$. Terms of the form $ijk$ with i,j,k all distinct appear 6 times each. We are interested in the sum of these terms each such term considered only once. Then, there are terms of the form $i^{2}jk$ with

$i \neq j$. Each such product appears three times. Finally, there are terms of  the form $i^{3}$. Each such term appears once only. Hence,

$A=6S_{10}+3B+C$ equation I

where $A=(\sum_{i=1}^{10})^{3}$ and $C=\sum_{i=1}^{i=10}i^{3}$ and

$\sum_{i \neq j}^{10}i^{2}j$.

Using the well-known formulae given below:

$1+2+3+ \ldots +n = \frac {n(n+1)}{2}$ equation II

$1^{2}+2^{2}+3^{2}+ \ldots + n^{3}= \frac {n^{2}(n+1)^{2}}{4}$ equation III

A and C come out to be respectively $(55)^{3}$ and $55^{2}$.

To calculate B, note that if we add to B, products of the form $i^{2}i$, then we get all the terms of the product

$(1^{2}+2^{2}+3^{2}+ \ldots + n^{2})(1+2+3+\ldots +n)$.

The first factor equals $55 \times 7$ using the formula

$\sum n^{2}=\frac {n(n+1)(2n+1)}{6}$ equation IV

Hence, $B=6 \times (55)^{2}$. Putting these values in equation I, we get

$S_{10}=(1/6)(A-3B-C)=18150$.

More later

Nalin Pithwa

### Factorization uses and tricks

One of the first concepts of elementary algebra is factorization. Apart from its intrinsic use in Math itself, it is applied in several areas like pole-placement method in Control System Design and design of digital filters for use in digital communications, the latter being the foundation stone of IT. Another use is in error control coding which is used in wireless/mobile/cellular communications. One of the trickiest uses of factorization is in Trigonometry, which of course, has countless applications. Let me present to you some tricks, which are not so obvious in factorizing complicated algebraic expressions.

1) Factorization by guessing! Solve the equation $x^{3}-3x^{2}-6x+16=0$.

Solution: The product of the coefficient of highest degree term and constant is 16. So, think of the factors of 16. You will see that the LHS vanishes when $x=2$. (We have used the factor theorem here).

Hence, $x=2$ is one root of the equation and corresponding to this root, we have a factor $x-2$.

The equation may now be written $x^{2}(x-2)-x(x-2)-8(x-2)=0$ or $(x^{2}-x-8)(x-2)=0$

Removing the factor $x-2$, we have $x^{2}-x-8=0$

hence, the three roots of the above cubic $x = 0.5(1+\sqrt{33}$ or $x = 0.5(1-\sqrt{33})$ or $x=2$.

Moral of the story: Guessing the factor reduces the degree of the equation. There will be some trials and errors though.

2) The Method of Undetermined Coefficients:

Resolve into two factors: $2a^{2}x^{2} - 2(3b-4c)(b-c)y^{2}+abxy$

Solution: This equation might seem not factorizable, at first sight, as the *middle term* involves no *c*. (Note: a, b, and c are parameters, and the variables are x and y). Also, the brute force technique for using the formula method for quadratics is way too cumbersome. So, let’s give a shot to it using the method of undetermined coefficients.

Let $2a^{2}x^{2} - 2(3b-4c)(b-c)y^{2}+abxy=(Px+Qy)(Mx+Ny)$

Hence, equating like coefficients, $PM=2a^{2}$, $NQ= -2(3b-4c)(b-c)$, $MQ+NP=ab$.

Again, you have to guess!

Let’s try: let $N=2(b-c)$, $Q=-(3b-4c)$, $M=a$, $P=2a$

We need to check: $MQ+NP=-a(3b-4c)+2(b-c)(2a)=ab$ Bingo:-)

So, the factors are $(ax+2(b-c)y)(2ax-y(3b-4c))$

3) An important factorization identity is:

$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

Would you like to try proving this?(It is not so easy!)

More later, — Nalin