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 –
DISCRIMINANTNATURE OF ROOTS
when b2-4ac < 0Roots are Imaginary and Unequal, Conjugate Pairs [(a+ib)(a-ib)]
when b2-4ac = 0Real 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 PRODUCTSIGN OF SUMSIGN OF THE ROOT
(+) positive(+) positive Both roots are Positive
(+) positive(-) negativeBooth roots are Negative
(-) negative(+) positiveNumerically larger root is Positive and the other root is Negative
(-) negative(-) negativeNumerically 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 anxn + an-1xn-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 $$g(-3) = (a-2)(-27) + (a+1)(9) + (-3) – 2a = -20a + 60 \\ g(-3) = 0 \Longrightarrow -20a + 60 = 0 \Longrightarrow a =3$$

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