CAT Quant – Simple and Quadratic Equations – Important Formulas and Concepts

Basics

Constants – fixed values [3, 1002, -9, 5/2]
Variables – does not have fixed value [x, y, t, z]
Expression – combination of constants and variables
Equation – when an expression = 0
Linear Equation in 1 Variable – (2x+3 = 0)
Linear Equation in 2 Variable – (3x+4y = 2)
Quadratic Equation – (2x²+7x+9 = 0)
Cubic Equation – (x+1 = x³)

Standard Equation

Given there are two Linear Equations in 2 variables $$a_1x+b_1y+c_1 = 0 \ , \ a_2x+b_2y+c_2 = 0$$
$${If} \ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \longrightarrow Infinite \ Solutions \ [Consistent]$$
$${If} \ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \longrightarrow No \ Solutions \ [Inconsistent]$$
$${If} \ \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \longrightarrow Unique \ Solutions \ [Consistent]$$

Digit Based Problems

Numbers with different number of digits like 2-digit, 3-digit and more can be expressed in the form of equations. Further, Digit based questions can be solved using Equations.
- 2-digit –> ab = (10a+b)
- 3-digit –> abc = (100a+10b+c)
- 4-digit –> abcd = (1000a+100b+10c+d)
Sum – Questions based on sum of a number and its reverse – $$”ab” + “ba” = [(10a+b)+(10b+a)] = 11(a+b) \longrightarrow Sum \ of \ digits \ is \ a \ multiple \ of \ 11 \ $$ $$”abcd” + “dcba” = [1001(a+d)+110(b+c)] \longrightarrow Sum \ of \ digits \ is \ a \ multiple \ of \ 11 \ $$ Note – Sum of Even Digits = Multiple of 11.
Difference – Questions based on difference of a number and its reverse – $$”ab” – “ba” = [(10a+b) – (10b+a)] = 9(a-b) \ or \ 9(b-a) \longrightarrow Difference \ of \ digits \ is \ a \ multiple \ of \ 9 \ $$ $$”abc” – “cba” = [99(a-c) \ or \ 99(c-a)] \longrightarrow Difference \ of \ digits \ is \ a \ multiple \ of \ 99 \ $$ $$”abcd” – “dcba” = [999(a-d) + 90(b-c)] \longrightarrow Difference \ of \ digits \ is \ a \ multiple \ of \ 9 \ $$ Note – for even difference = Multiple of 9 ; for odd difference = Multiple of 99

Theory of Equation –

Some basic rules pertaining to equations of different types are as follows –
- An equation of degree “n” has at most “n” real roots. For Example – an equation with degree 2 (11x²+3x+9 =0) will have 2 roots.
- Total number of Roots = Number of Real Roots + Number of Imaginary Roots
- Imaginary Roots always occur in pairs [Conjugate pairs (a+ib , a-ib)]

Quadratic Equation

An equation of the form –> (ax²+bx+c = 0) where a, b, c are all Real and a not equals 0 is called a quadratic equation.
Highest degree of “n” = 2
Number of Roots / Solution = 2 (real or imaginary)

Methods of finding Roots

In general, the roots of a quadratic equation can be found out in 2 ways –
- Factorizing the expression
- Discriminant (Standard formula)

Factorizing Method

A quadratic equation ax²+bx+c = 0 can be written in the form (x-α)(x-β)=0 where α and β are the roots. The method of factorizing the expression on the left hand side of the quadratic equation is explained with an example — x²+7x+12 = 0.
- 1st Step – Write down b (the coefficient of x) as the sum of two quantities whose product is equal to ac. In this case “7x” can be written as the sum of (-3x) + (-4x)so that the product is equal to +12.
- 2nd Step – Rewrite the equation with the “bx” term split in the above manner. In this case, the above equation can be written as x²-3x-4x+12 = 0.
- 3rd Step – Take the first two term and rewrite them together after taking out the common factor between them and then do the same with the third and the fourth term. You should ensure that the term left after taking out the common factor from the first and second term is the same as that of the third and fourth term. In this case, the equation can be rewritten as x(x-3)-4(x-3) = 0.
- 4th Step – Re write the entire left hand side to get the form (x-α)(x-β)=0. In this case, we can rewrite the given equation as (x-3)(x-4) = 0.
- 5th Step – Now, α and β are the roots of the quadratic equation. For x²+7x+12 = 0, the roots of the equation are 3 and 4.

Discriminant Method

For a quadratic equation of the form ax²+bx+c = 0 ; the standard formula for finding the root is given by – $$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \\ where, \ Discriminant \ (D) \ = \sqrt{b^2-4ac}$$
Certain rules regarding roots of a quadratic equation equation –
- Sum of Roots = -b/a
- Product of Roots = c/a

Nature of the Roots

The discriminant determines the nature of roots in a quadratic equation. The natures of roots when a, b, c are rational are –

DISCRIMINANT	NATURE OF ROOTS
when b²-4ac < 0	Roots are Imaginary and Unequal, Conjugate Pairs [(a+ib)(a-ib)]
when b²-4ac = 0	Real and Equal Roots
when b²-4ac > 0 [Perfect Square]	Real and Distinct Roots (Rational)
when b²-4ac > 0 [not Perfect Square]	Irrational and Distinct Roots, Conjugate Pairs $ [(a + \sqrt{b})(a – \sqrt{b})] $

Sign of the Roots

The relationship between the sum and the product of the roots and the signs of the roots are as follows –

SIGN OF PRODUCT	SIGN OF SUM	SIGN OF THE ROOT
(+) positive	(+) positive	Both roots are Positive
(+) positive	(-) negative	Booth roots are Negative
(-) negative	(+) positive	Numerically larger root is Positive and the other root is Negative
(-) negative	(-) negative	Numerically larger root is Negative and the other root is Positive

Maximum and Minimum Value of a Quadratic Equation

The quadratic expression and take different values as x takes different values. As x varies from $ -\infty \ to \ +\infty $
Graph of a Quadratic Equation is always a Parabola.
f(x) = y = ax²+bx+c where a, b, c are real and a $ \neq $ 0 then y = f(x) represents the parabola .

Case I : Minimum Value [a>0]

f(x) has a minimum value whenever a>0 [always positive] $$Point \ of \ Minima = \frac{-b}{2a} \ \quad \ Value \ of \ Minima = {\LARGE [} \frac{-D}{4a} {\LARGE ]}{\LARGE [} \frac{x+β}{2}{\LARGE ]}$$
When, a>0 and D>0 –
- Real and Distinct Roots
- Minimum Value – f(x) will be (+)positive for all Real Values of “x” which is when x does not lie between α and β [(-)negative value between α and β]
When, a>0 and D<0 –
- Roots are Imaginary
- Minimum Value – f(x) will be (+)positive for all values of “x”
When, a>0 and D=0 –
- Equal Roots
- Minimum Value – f(x) will be (+)positive for all values of “x” except when f(x) = 0.

Case II : Maximum Value [a<0]

f(x) has a maximum value whenever a<0 [always negative] $$Point \ of \ Minima = \frac{-b}{2a} \ \quad \ Value \ of \ Minima = {\LARGE [} \frac{-D}{4a} {\LARGE ]}{\LARGE [} \frac{x+β}{2}{\LARGE ]}$$
When, a<0 and D>0 –
- Real and Distinct Roots
- Minimum Value – f(x) will be (-)negative for all Real Values of “x” which is when x does not lie between α and β [(+)positive value between α and β]
When, a<0 and D<0 –
- Roots are Imaginary
- Minimum Value – f(x) will be (-)negative for all values of “x”
When, a>0 and D=0 –
- Equal Roots
- Minimum Value – f(x) will be (-)negative for all values of “x” except when f(x) = 0.

Polynomial

Polynomial Equation

An expression of the type a_nx_n + a_n-1x^n-1+ ….. + a₁x a₀ is called a Polynomial. If we denote it as f(x), then f(x) = 0 is a Polynomial Equation.

Remainder Theorem

A polynomial p(x) of degree n is divided by x-a (a linear polynomial), there results a quotient polynomial q(x) of degree (n-1) and a remainder of degree 0 i.e a quotient.$$p(x) = (x-a) * q(x) + R$$ This is true for all values of x. If x=a –> p(a) = R
If p(a) = 0, we say that “a” is a zero of the polynomial p(x)
If p(x) is a polynomial and “a” is a zero of p(x) then p(x) = (x-a)q(x).
If p(x) is divided by (ax+b), then the remainder is given by p(-b/a).
If p(x) is divided by (ax-b), then the remainder is given by p(b/a).
The degree of remainder is always less then degree of divisor.

Factor Theorem

If R=0, i.e p(a) = 0, then (x-a) is a factor of p(x) then we have p(a) = 0. This immediate consequence of the Remainder Theorem is called the Factor Theorem
The number “a” is a root of p(x)=0 if and only if (x-a) is a factor of p(x).
Example – if (a-2)x³+(a-1)x²+x-2a is divisible by x+3, find the value of a.
- Let, g(x) = (a-2)x³+(a-1)x²+x-2a. Here according to factor theorem x-a is a factor of f(x) if and only if f(a)= 0. Also g(x) is divisible by (x+3) i.e [x-(-3)] is a factor of g(x). Thus g(-3) = 0 $$g(-3) = (a-2)(-27) + (a+1)(9) + (-3) – 2a = -20a + 60 \\ g(-3) = 0 \Longrightarrow -20a + 60 = 0 \Longrightarrow a =3$$