Simplified Quantitative Formulas: Equations
- Equation: A statement that two expressions are equal. E.g., 2x+3=7.
- Linear: ax+b=0. Solution: x=-b/a.
- Quadratic: ax²+bx+c=0. Solution: x=[-b±√(b²-4ac)]/(2a).
- Simultaneous: Two or more equations solved together.
- Cubic/Polynomial: Higher degree equations.
- Key Properties: Roots, discriminant, sum/product of roots.
What do these mean? (Super Simple Explanations & Examples)
- Equation: Like a balance: both sides must be equal.
- Example: 2x+3=7. To solve: 2x=4, x=2. - Linear: Only x, no x². Example: x+5=9 ⇒ x=4.
- Quadratic: Has x². Example: x²-4=0 ⇒ x=2 or -2.
- Simultaneous: Solve two equations together. Example: x+y=5, x-y=1 ⇒ x=3, y=2.
- Cubic: Has x³. Example: x³-8=0 ⇒ x=2.
Linear Equations
Linear equations are equations of the first degree, where the highest power of the variable is 1.
General Form of Linear Equations
A linear equation is an equation of the first degree, meaning the highest power of the variable is 1. It can be written in the form ax + b = 0, where a and b are constants and a ≠ 0.
General form: ax + b = 0, where a ≠ 0
Solution: x = -b/a
Example: 2x + 5 = 0
Here, a = 2, b = 5
Solution: x = -5/2 = -2.5
- Linear equations are the backbone of Quant. Their concepts often appear in other topics too.
- Graphically, a linear equation in two variables (ax + by = c) represents a straight line. The solution to two such equations is the intersection point of their lines.
- For two equations:
- If the ratios of coefficients of x and y are equal, but the constant ratio is different, the lines are parallel (no solution).
- If all ratios are equal, the lines coincide (infinite solutions).
- If the ratios of x and y are different, the lines intersect at a unique point (one solution).
- For n variables, you generally need n independent equations for a unique solution. Fewer equations usually mean infinite or indeterminate solutions, unless extra conditions (like integer values) are given.
- For three variables, the system can be solved using determinants:
- If the main determinant (D) ≠ 0, there is a unique solution.
- If D = 0, but at least one of Dx, Dy, Dz is nonzero, there is no solution.
- If D = Dx = Dy = Dz = 0, there are infinitely many solutions.
- CAT Tip: Use elimination, substitution, or LCM methods for speed. For two equations, align coefficients and subtract to eliminate variables quickly.
Example 1: Basic Linear Equations
Q1.
Solve the equation: 3x + 5 = 20
Solution:
3x + 5 = 20
3x = 20 - 5
3x = 15
x = 15/3 = 5
Q2.
Solve the equation: 2(x + 3) = 4x - 6
Solution:
2(x + 3) = 4x - 6
2x + 6 = 4x - 6
2x - 4x = -6 - 6
-2x = -12
x = 6
Q3.
Solve the equation: (x + 2)/3 = (2x - 1)/4
Solution:
(x + 2)/3 = (2x - 1)/4
4(x + 2) = 3(2x - 1)
4x + 8 = 6x - 3
4x - 6x = -3 - 8
-2x = -11
x = 11/2
Q4.
Solve the equation: 5x - 3 = 2x + 9
Solution:
5x - 3 = 2x + 9
5x - 2x = 9 + 3
3x = 12
x = 4
Quadratic Equations
Quadratic equations are equations of the second degree, where the highest power of the variable is 2.
Quadratic Equations
A quadratic equation is an equation of the second degree, where the highest power of the variable is 2. It can be written in the form ax² + bx + c = 0, where a, b, and c are constants and a ≠ 0.
General form: ax² + bx + c = 0, where a ≠ 0
Quadratic formula: x = (-b ± √(b² - 4ac))/2a
Discriminant: D = b² - 4ac
Example: x² + 5x + 6 = 0
Here, a = 1, b = 5, c = 6
D = 25 - 24 = 1 > 0 (two real roots)
x = (-5 ± √1)/2 = -2 or -3
- Quadratic equations are crucial for CAT. Mastery of their properties can save time in the exam.
- The roots of ax² + bx + c = 0 are given by x = [-b ± √(b²-4ac)]/(2a). The discriminant (D = b²-4ac) determines the nature of the roots:
- D > 0 and a perfect square: roots are real, rational, and distinct.
- D > 0 and not a perfect square: roots are real, irrational, and distinct.
- D = 0: roots are real and equal.
- D < 0: roots are complex conjugates.
- Sum of roots = -b/a, Product of roots = c/a.
- Signs of roots:
- If product > 0 and sum > 0: both roots positive.
- If product > 0 and sum < 0: both roots negative.
- If product < 0: roots have opposite signs.
- Minimum/maximum value of ax² + bx + c occurs at x = -b/(2a).
- If a > 0: minimum value = (4ac-b²)/(4a).
- If a < 0: maximum value = (4ac-b²)/(4a).
- CAT Tip: Use options, value plugging, and symmetry for quick solutions. If roots are reciprocals, c = a. If b = 0, roots are equal in magnitude and opposite in sign.
- Descartes' Rule: The number of sign changes in the polynomial gives the maximum number of positive roots. For negative roots, check f(-x).
- Any polynomial of odd degree always has at least one real root.
Example 1: Solving Quadratic Equations
Q1.
Solve the equation: x² + 5x + 6 = 0
Solution:
Using factorization:
x² + 5x + 6 = 0
(x + 2)(x + 3) = 0
x = -2 or x = -3
Q2.
Solve the equation: 2x² - 7x + 3 = 0
Solution:
Using quadratic formula:
a = 2, b = -7, c = 3
x = (7 ± √(49 - 24))/4
x = (7 ± √25)/4
x = (7 ± 5)/4
x = 3 or x = 1/2
Q3.
Solve the equation: x² - 4x + 4 = 0
Solution:
Using perfect square:
x² - 4x + 4 = 0
(x - 2)² = 0
x = 2 (repeated root)
Q4.
Solve the equation: x² + 2x + 5 = 0
Solution:
Using quadratic formula:
a = 1, b = 2, c = 5
x = (-2 ± √(4 - 20))/2
x = (-2 ± √(-16))/2
x = (-2 ± 4i)/2
x = -1 ± 2i (complex roots)
Simultaneous Equations
Simultaneous equations are a set of equations containing multiple variables.
Methods for Solving Simultaneous Equations
Simultaneous equations are a set of equations containing multiple variables. There are several methods to solve them:
1. Substitution Method: Solve one equation for one variable and substitute into other equations.
2. Elimination Method: Add or subtract equations to eliminate one variable.
3. Cross-multiplication Method: Use determinants to solve two linear equations.
Example: 2x + y = 5 and 3x - 2y = 4
Using elimination:
Multiply first by 2: 4x + 2y = 10
Add to second: 7x = 14
x = 2, y = 1
Example 1: Linear Simultaneous Equations
Q1.
Solve the equations: 2x + y = 5 and 3x - 2y = 4
Solution:
Using elimination method:
Multiply first equation by 2: 4x + 2y = 10
Add to second equation: 7x = 14
x = 2
Substitute in first equation: 2(2) + y = 5
y = 1
Q2.
Solve the equations: x + 2y = 7 and 2x - y = 4
Solution:
Using substitution method:
From first equation: x = 7 - 2y
Substitute in second equation: 2(7 - 2y) - y = 4
14 - 4y - y = 4
-5y = -10
y = 2
x = 7 - 2(2) = 3
Q3.
Solve the equations: 3x + 4y = 10 and 2x + 3y = 7
Solution:
Using cross-multiplication:
x/(4×7 - 3×10) = y/(10×2 - 7×3) = 1/(3×3 - 2×4)
x/(-2) = y/(-1) = 1/1
x = 2, y = 1
Q4.
Solve the equations: 5x - 3y = 11 and 2x + 5y = 1
Solution:
Using elimination method:
Multiply first by 2: 10x - 6y = 22
Multiply second by 5: 10x + 25y = 5
Subtract: -31y = 17
y = -17/31
Substitute to find x = 58/31
Cubic Equations
Cubic equations are equations of the third degree, where the highest power of the variable is 3.
Cubic Equations
A cubic equation is an equation of the third degree, where the highest power of the variable is 3. It can be written in the form ax³ + bx² + cx + d = 0, where a, b, c, and d are constants and a ≠ 0.
General form: ax³ + bx² + cx + d = 0, where a ≠ 0
Methods of solving:
1. Factorization: Find common factors or use synthetic division
2. Cardano's formula: For general cubic equations
3. Synthetic division: To find rational roots
Example: x³ - 6x² + 11x - 6 = 0
Using factorization:
(x - 1)(x - 2)(x - 3) = 0
x = 1, 2, or 3
Example 1: Solving Cubic Equations
Q1.
Solve the equation: x³ - 6x² + 11x - 6 = 0
Solution:
Using factorization:
x³ - 6x² + 11x - 6 = 0
(x - 1)(x - 2)(x - 3) = 0
x = 1, 2, or 3
Q2.
Solve the equation: x³ + 3x² - 4x - 12 = 0
Solution:
Using synthetic division:
Try x = 2:
2 | 1 3 -4 -12
2 10 12
1 5 6 0
x = 2 is a root
Factor: (x - 2)(x² + 5x + 6) = 0
x = 2, -2, or -3
Q3.
Solve the equation: x³ - 3x + 2 = 0
Solution:
Using Cardano's formula:
Let x = y - b/(3a)
y³ + py + q = 0
p = -3, q = 2
Using formula: x = 1, 1, -2
Q4.
Solve the equation: x³ + 2x² - 5x - 6 = 0
Solution:
Using rational root theorem:
Possible roots: ±1, ±2, ±3, ±6
Try x = 2:
2³ + 2(2)² - 5(2) - 6 = 0
x = 2 is a root
Factor: (x - 2)(x² + 4x + 3) = 0
x = 2, -1, or -3
Polynomial Equations
Polynomial equations are equations containing terms with different powers of the variable.
Polynomial Equations
A polynomial equation is an equation containing terms with different powers of the variable. The degree of the polynomial is the highest power of the variable.
General form: aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀ = 0
Important theorems:
1. Factor theorem: If f(a) = 0, then (x - a) is a factor of f(x)
2. Remainder theorem: The remainder when f(x) is divided by (x - a) is f(a)
3. Rational root theorem: Possible rational roots are ±p/q, where p divides a₀ and q divides aₙ
Example: x⁴ - 5x² + 4 = 0
Let y = x²
y² - 5y + 4 = 0
(y - 1)(y - 4) = 0
x = ±1 or ±2
Example 1: Solving Polynomial Equations
Q1.
Solve the equation: x⁴ - 5x² + 4 = 0
Solution:
Let y = x²
y² - 5y + 4 = 0
(y - 1)(y - 4) = 0
y = 1 or 4
x = ±1 or ±2
Q2.
Solve the equation: x⁴ + 2x³ - 7x² - 8x + 12 = 0
Solution:
Using rational root theorem:
Possible roots: ±1, ±2, ±3, ±4, ±6, ±12
Try x = 1: 1 + 2 - 7 - 8 + 12 = 0
x = 1 is a root
Factor: (x - 1)(x³ + 3x² - 4x - 12) = 0
x = 1, 2, -2, or -3
Q3.
Solve the equation: x⁵ - 3x⁴ + 2x³ = 0
Solution:
Factor out x³:
x³(x² - 3x + 2) = 0
x³(x - 1)(x - 2) = 0
x = 0 (triple root), 1, or 2
Q4.
Solve the equation: x⁴ - 16 = 0
Solution:
Using difference of squares:
(x² - 4)(x² + 4) = 0
(x - 2)(x + 2)(x² + 4) = 0
x = 2, -2, or ±2i
Advanced Concepts
Advanced topics in equations including complex numbers, inequalities, and applications.
Advanced Concepts
Advanced topics in equations include complex numbers, inequalities, and absolute values. These concepts are essential for solving more complex mathematical problems.
Complex numbers: a + bi, where i² = -1
Inequalities: >, <, ≥, ≤
Absolute value: |x| = x if x ≥ 0, -x if x < 0
Example: x² + 4 = 0
x² = -4
x = ±√(-4)
x = ±2i (complex roots)
Example 1: Complex Numbers
Q1.
Solve the equation: x² + 4 = 0
Solution:
x² = -4
x = ±√(-4)
x = ±2i
Q2.
Solve the equation: x² - 2x + 5 = 0
Solution:
Using quadratic formula:
x = (2 ± √(4 - 20))/2
x = (2 ± √(-16))/2
x = 1 ± 2i
Q3.
Solve the inequality: x² - 4x + 3 > 0
Solution:
Factor: (x - 1)(x - 3) > 0
Critical points: x = 1, 3
Test intervals: x < 1, 1 < x < 3, x > 3
Solution: x < 1 or x > 3
Q4.
Solve the equation: |2x - 3| = 5
Solution:
2x - 3 = 5 or 2x - 3 = -5
2x = 8 or 2x = -2
x = 4 or x = -1
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