Equations and matrices– CA Foundation, CPT notes, PDF
This article is about Equations and matrices – Business Mathematics and Logical Reasoning & Statistics for CA foundation CPT students. we also provide PDF file at the end.
What we will study in this chapter: Equations and matrices
LEARNING OBJECTIVES
After studying this chapter, you will be able to:
Understand the concept of equations and its various degrees – linear, simultaneous, quadratic and cubic equations.
Know how to solve the differential equations using different methods of solutions.
UNIT OVERVIEW
2.1.1 INTRODUCTION:
The equation is defined to be a mathematical statement of equality. If equality is true for a certain value of the variable involved, the equation is often called a conditional equation and equality sign ‘=’ is used; while if the equality is true for all values of the variable involved, the equation is called an identity.
For Example: \frac { x+2 }{ 3 } +\frac { x+3 }{ 2 } =3 holds true only for
So it is a conditional. On the other hand,
is an identity since it holds for all values of the variable x.
Determination of value of the variable which satisfies an equation is called solution of the equation or root of the equation. An equation in which highest power of the variable is 1 is called a Linear (or a simple) equation. This is also called the equation of degree 1. Two or more linear equations involving two or more variables are called Simultaneous Linear Equations. An equation of degree 2 (highest Power of the variable is 2) is called Quadratic equation and the equation of degree 3 is called Cubic Equation.
For Example: is a Linear equation.
is a Quadratic equation.
is a Cubic equation.
are jointly called Simultaneous equations.
2.1.2 SIMPLE EQUATION
A simple equation in one unknown x is in the form
Where a, b are known constants and
Note: A simple equation has only one root.
Example:
Solution: By transposing the variables in one side and the constants in other side we have
UNIT-I : EXERCISE (A)
Choose the most appropriate option (a) (b) (c) or (d).
1. | The equation will be satisfied for x equal to: a) 2 b) -1 c) 1 d) none of these |
2. | The root of the equation is a) 20 b) 10 c) 2 d) none of these |
3. | Pick up the correct value of x for a) b) c) d) none of these |
4. | The solution of the equation a) b) c) d) none of these |
5. | 8 is the solution of the equation a) b) c) d) |
6. | The value of y that satisfies the equation is a) b) c) d) |
7. | The solution of the equation is a) b) c) d) none of these |
8. | The equation is true for a) b) c) d) |
9. | Pick up the correct value x for which a) b) c) d) none of these |
ILLUSTRATIONS:
1. | The denominator of a fraction exceeds the numerator by 5 and if 3 be added to both the fraction becomes. Find the fraction. Let x be the numerator and the fraction be By the question or or The required fraction is |
2. | If thrice of A’s age 6 years ago be subtracted from twice his present age, the result would be equal to his present age. Find A’s present age. Let x years be A’s present age. By the question Or Or Or Or A’s present age is 9 years. |
3. | A number consists of two digits the digit in the ten’s place is twice the digit in the unit’s place. If 18 be subtracted from the number the digits are reversed. Find the number. Let x be the digit in the unit’s place. So the digit in the ten’s place is 2x. Thus the number becomes By the question Or Or Or So the required number is |
4. | For a certain commodity the demand equation giving demand ‘d’ in kg, for a price ‘p’ in rupees per kg. isThe supply equation giving the supply in kg. for a price p in rupees per kg. isThe market price is such at which demand equals supply. Find the market price and quantity that will be bought and sold. Given and Since the market price is such that demand (d) = supply (s) we have or Or So market price of the commodity is ₹7 per kg. the required quantity bought kg. And the quantity sold kg. |
UNIT-I : EXERCISE
(B) Choose the most appropriate option (a) (b) (c) or (d).
1. | The sum of two numbers is 52 and their difference is 2. The numbers are a) 17 and 15 b) 12 and 10 c) 27 and 25 d) none of these
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2. | The diagonal of a rectangle is 5 cm and one of at sides is 4 cm. Its area is a) 20 sq.cm. b) 12 sq.cm. c) 10 sq.cm. d) none of these
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3. | Divide 56 into two parts such that three times the first part exceeds one third of the second by 48. The parts are. a) (20, 36) b) (25, 31) c) (24, 32) d) none of these
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4. | The sum of the digits of a two digit number is 10. If 18 be subtracted from it the digits in the resulting number will be equal. The number is a) 37 b) 73 c) 75 d) none of these
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5. | The fourth part of a number exceeds the sixth part by 4. The number is a) 84 b) 44 c) 48d) none of these
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6. | Ten years ago the age of a father was four times of his son. Ten years hence the age of the father will be twice that of his son. The present ages of the father and the son are. a) (50, 20) b) (60, 20) c) (55, 25) d) none of these
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7. | The product of two numbers is 3200 and the quotient when the larger number is divided by the smaller is 2.The numbers are a) (16, 200) b) (160, 20) c) (60, 30) d) (80, 40)
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8. | The denominator of a fraction exceeds the numerator by 2. If 5 be added to the numerator the fraction increases by unity. The fraction is. a) b) c) d)
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9. | Three persons Mr. Roy, Mr. Paul and Mr. Singh together have ₹ 51. Mr. Paul has ₹ 4 less than Mr. Roy and Mr. Singh has got ₹ 5 less than Mr. Roy. They have the money as.
a) (₹ 20, ₹16, ₹ 15) b) (₹ 15, ₹20, ₹ 16) c) (₹ 25, ₹11, ₹ 15) d) none of these
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10 | A number consists of two digits. The digits in the ten’s place is 3 times the digit in the unit’s place. If 54 is subtracted from the number the digits are reversed. The number is
a) 39 b) 92 c) 93 d) 94
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11 | One student is asked to divide a half of a number by 6 and other half by 4 and then to add the two quantities. Instead of doing so the student divides the given number by 5. If the answer is 4 short of the correct answer then the number was a) 320 b) 400c) 480 d) none of these.
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12 | If a number of which the half is greater than th of the number by 15 then the number is a) 50 b) 40 c) 80 d) none of these. |
2.1.3 SIMULTANEOUS LINEAR EQUATIONS IN TWO UNKNOWNS
The general form of a linear equations in two unknowns x and y is where a, b are non-zero coefficients and c is a constant. Two such equations and form a pair of simultaneous equations in x and y. A value for each unknown which satisfies simultaneously both the equations will give the roots of the equations.
2.1.4 METHOD OF SOLUTION
Elimination Method: In this method two given linear equations are reduced to a linear equation in one unknown by eliminating one of the unknowns and then solving for the other unknown.
Example 1: Solve: and
Solution:
By making
and by making
Adding. Substituting this values of x in we find;
Cross Multiplication Method: Let two equations be:
We write the coefficients of x, y and constant terms and two more columns by repeating the coefficients of x and y as follows:
1 2 3 4 b1 c1 a1 b1 b2 c2 a2 b2
and the result is given by:
so the solution is :
Example 2: Solve
Solution:
Method of elimination: By we get
Adding (ii) & (iii) we get
or
Putting we get
or,
So
Method of cross-multiplication:
or
or
or
2.1.5 METHOD OF SOLVING SIMULTANEOUS LINEAR
EQUATION WITH THREE VARIABLES
Example 1: Solve for x, y and z:
Solution: (a) Method of elimination
By (i) × 2 we get
By [the variable z is thus eliminated]
By
By
By
Putting
Putting
or
or
or
So is the required solution.
(b) Method of cross multiplication
We write the equations as follows:
By cross multiplication
Substituting above values for x and y in equation (iii) i.e., we have
[the variable z is thus eliminated]
or
or
Now
Thus
Example 2: Solve for x y and z:
Solution: We put and get
By
By
or
By
By
By
Putting
or By
or By
Putting
or
Thus is the solution.
Example 3: Solve for x, y and z
Solution: We can write as
By we get
or
By
By
Required solution is
UNIT-I : EXERCISE (C)
Choose the most appropriate option (a), (b), (c) or (d).
1. The solution of the set of equations is a) b) c) d)
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2. The values of x and y satisfying the equations are given by the pair. a) b) c) d) none of these |
3. are satisfied by the values given by the pair. a) b) c) d) none of these |
4. The solution for the pair of equations is given by a) b) c) d)
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5. Solve for x and and. a) b) c) d)
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6. The pair satisfying the equations is given by a) b) c) d) none of these
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7. Solve for x and a) b) c) d) none of these
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8. The simultaneous equations have solutions given by a) b) c) d)
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9. have solutions as a) b) c) d)
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10. The values of x and y satisfying the equations are given by a) b) c) d) |
UNIT-I : EXERCISE (D)
Choose the most appropriate option (a), (b), (c) or (d) as the solution to the given set of equations:
1. a) b) c) d)
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2. a) b) c) d) |
3. a) b) c) d)
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4. a) b) c) d)
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5. a) b) c) d)
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6. a) b) c) d)
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7. a) b) c) d)
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8. a) b) c) d)
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9. a) b) c) d)
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10. a) b) c) d) |
2.1.6 PROBLEMS LEADING TO SIMULTANEOUS EQUATIONS
ILLUSTRATIONS:
1. | If the numerator of a fraction is increased by 2 and the denominator by 1 it becomes 1. Again if the numerator is decreased by 4 and the denominator by 2 it becomes 1/2 . Find the fraction. |
SOLUTION: Let be the required fraction. By the question Thus and By From So the required fraction is
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2. | The age of a man is three times the sum of the ages of his two sons and 5 years hence his age will be double the sum of their ages. Find the present age of the man? SOLUTION: Let x years be the present age of the man and sum of the present ages of the two sons be y years. By the condition And From (i) & (ii) or or or Hence the present age of the main is 45 years
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3. | A number consist of three digit of which the middle one is zero and the sum of the other digits is 9. The number formed by interchanging the first and third digits is more than the original number by 297 find the number. SOLUTION: Let the number be we have Also From or Adding (i) and (ii) . from Hence the number is 306. |
UNIT-I : EXERCISE (E)
Choose the most appropriate option (a), (b), (c) or (d).
1. | Monthly incomes of two persons are in the ratio 4 : 5 and their monthly expenses are in the ratio 7 : 9. If each saves ₹ 50 per month find their monthly incomes. a) b) c) d)
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2. | Find the fraction which is equal to 1/2 when both its numerator and denominator are increased by 2. It is equal to 3/4 when both are increased by 12. a) b) c) d)
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3. | The age of a person is twice the sum of the ages of his two sons and five years ago his age was thrice the sum of their ages. Find his present age. a) b) c) d)
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4. | A number between 10 and 100 is five times the sum of its digits. If 9 be added to it the digits are reversed find the number. a) b) c) d)
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5. | The wages of 8 men and 6 boys amount to ₹33. If 4 men earn ₹4.50 more than 5 boys determine the wages of each man and boy. a) (₹ 1.50, ₹3) b) (₹ 3, ₹ 1.50) c) (₹ 2.50, ₹ 2) d) (₹ 2, ₹ 2.50)
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6. | A number consisting of two digits is four times the sum of its digits and if 27 be added to it the digits are reversed. The number is a) b) c) d)
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7. | Of two numbers, 1/5th of the greater is equal to 1/3rd of the smaller and their sum is 16. The numbers are: a) b) c) d)
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8. | y is older than x by 7 years 15 years back x’s age was 3/4 of y’s age. Their present ages are: a) b) c) d)
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9. | The sum of the digits in a three digit number is 12. If the digits are reversed the number is increased by 495 but reversing only of the ten’s and unit digits increases the number by 36. The number is a) b) c) d)
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10. | Two numbers are such that twice the greater number exceeds twice the smaller one by 18 and 1/3rd of the smaller and 1/5th of the greater number are together 21. The numbers are: a) b) c) d)
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11. | The demand and supply equations for a certain commodity are and respectively where p is the market price and q is the quantity then the equilibrium price and quantity are: a) b) c) d) None of these |
2.1.7 QUADRATIC EQUATION
An equation of the form where x is a variable and a, b, c are constants with a is called a quadratic equation or equation of the second degree.
When the equation is called a pure quadratic equation; when the equation is called an affected quadratic.
Examples:
- i)
- ii)
iii)
The value of the variable say x is called the root of the equation. A quadratic equation has got two roots.
How to find out the roots of a quadratic equation:
or
or
or
or
or
Sum and Product of the Roots:
Let one root be and the other root be
Now
Thus sum of roots
Net
So the product of the roots
2.1.8 HOW TO CONSTRUCT A QUADRATIC EQUATION
For the equation we have
or
or
or
2.1.9 NATURE OF THE ROOTS
- i) If the roots are real and equal;
- ii) If then the roots are real and unequal (or distinct);
iii) If then the roots are imaginary;
- iv) Ifperfect squarethe roots are real, rational and unequal (distinct);
- v) If but not a perfect square the rots are real, irrational and unequal.
Since discriminates the roots is called the discriminate in the equationas it actually discriminates between the roots.
Note: (a) Irrational roots occur in conjugate pairs that is if is a root then is the other root of the same equation.
(b) If one root is reciprocal to the other root then their product is 1 and so
(c) If one root is equal to other root but opposite in sign then.
their sum = 0 and so i.e.
Example 1: Solve
Solution: 1st method :
or
or
or
or
2nd method (By formula)
Here a = 1, b = –5 , c = 6 (comparing the equation with
Example 2: Examine the nature of the roots of the following equations.
- i) ii)
iii) iv)
Solution: (i) a = 1, b = –8, c = 16
The roots are real and equal.
(ii)
and a perfect square
The roots are real, rational and unequal
(iii)
The roots are imaginary and unequal
(iv)
The roots are real and unequal. Since is not a perfect square the roots are realirrational and unequal.
ILLUSTRATIONS:
- If and be the roots of find the equation whose roots are
SOLUTION: Now sum of the roots of the required equation
2
Product of the roots of the required equation
Hence the required equation is
or
- If be the roots of find the value of
SOLUTION:
- Solve
SOLUTION:
or
or ( taking or )
or
or
orTherefore
- Solve
Solution:
or
or
or Taking
or
or
or
or
i.e.
i.e.
- Solve i.e.
SOLUTION
or
or
orwhen
or
or
or
or
or
For i.e,
For i.e,
- If one root of the equation is from the equation given that the roots are irrational
SOLUTION other roots is sum of two roots
Product of roots
Required equation is : – (sum of roots )x + (Product of roots) = 0
or
- If are the two of the equation from the equation
whose roots are
SOLUTION : As are the roots of the equation
Now
Required equation is
or
- If the roots of the equation
are equal show that
SOLUTION: Since the roots of the given equation are equal the discriminant must be
zeroie.
or
or
or
or
UNIT-I : EXERCISE (F)
Choose the most appropriate option (a) (b) (c) or (d).
1. If the roots of the equation are equal then value of m is a) b) c) d)
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2. If then values of x are a) b) c) d)
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3. The values of a) b) c) d) none of these
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4. If be the roots of the equation the value of is a) b) c) d)
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5. If the sum of the roots of the quadratyic equation is equal to the sum of the squares of their reciprocals then is equal to a) b) c) d)
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6. The equation has equal roots the values of p will be. a) b) c) d)
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7. The roots of the equationare real if. a) b) c) d)
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8. If is one of the solutions of the equation the possible values of m are a) b) c) d)
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9. If p and q are the roots of x2 + 2x + 1 = 0 then the values of p3 + q3 becomes a) b) c) d)
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10. If L + M + N = 0 and L, M, N are rationals the roots of the equation are a) real and irrational b) real and rational c) imaginary and equal d) real and equal
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11. If are the roots of then value of is a) b) c) d)
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12. If and and the equation having roots as and is a) b) c) d)
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13. If one root of be reciprocal of the other then the value of p is a) b) c) d) |
UNIT-I : EXERCISE (G)
Choose the most appropriate option (a) (b) (c) or (d).
1. A solution of the quadratic equation is a) b) c) d)
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2. If the root of the equation exceeds the other by 4 then the value of m is a) b) c) d)
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3. The values of x in the equation are a) b) c) d)
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4. The solutions of the equationare a) b) c) d)
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5. The satisfying values of x for the equation are a) b) c) d)
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6. The values of x for the equation are a) b) c) d)
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7. The values of x satisfying the equation are a) b) c) d)
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8. The solution of the equation are a) b) c) d)
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9. The equation has got the solution as a) b) c) d)
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10. The equation has got two values of x to satisfy the equation given as a) b) c) d) |
2.1.10 PROBLEMS ON QUADRATIC EQUATION
1 | Difference between a number and its positive square root is 12; find the numbers? Solution: Let the number be x. Then Taking Or Either i.e, Either Ifif does not satisfy equation (i) so
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2. | A piece of iron rod costs ₹ 60. If the rod was 2 metre shorter and each metre costs ₹ 1.00 more, the cost would remain unchanged. What is the length of the rod? Solution: Let the length of the rod be xmetres. The rate per meter is ₹ New Length as the cost remain the same the new rate per meter is As given Or Or Or Or or Either (not possible) Hence the required length = 12m.
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3. | Divide 25 into two parts so that sum of their reciprocals is 1/6. Solution: let the parts be x and By the question or or or or or or So the parts of 25 are 10 and 15.
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UNIT-I : EXERCISE (H)
Choose the most appropriate option (a) (b) (c) or (d).
1. | The sum of two numbers is 8 and the sum of their squares is 34. Taking one number as x form an equation in x and hence find the numbers. The numbers are a) b) c) d)
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2. | The difference of two positive integers is 3 and the sum of their squares is 89. Taking the smaller integer as x form a quadratic equation and solve it to find the integers. The integers are. a) b) c) d)
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3. | Five times of a positive whole number is 3 less than twice the square of the number. The number is a) b) c) d)
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4. | The area of a rectangular field is 2000 sq.m and its perimeter is 180m. Form a quadratic equation by taking the length of the field as x and solve it to find the length and breadth of the field. The length and breadth are a) b) c) d) none
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5. | Two squares have sides p cm and cms. The sum of their squares is 625 sq. cm. The sides of the squares are a) b) c) d) none of these
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6. | Divide 50 into two parts such that the sum of their reciprocals is 1/12. The numbers are a) b) c) d)
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7. | There are two consecutive numbers such that the difference of their reciprocals is 1/240. The numbers are a) b) c) d)
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8. | The hypotenuse of a right–angled triangle is 20cm. The difference between its other two sides be 4cm. The sides are a) b) c) d) none of these
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9. | The sum of two numbers is 45 and the mean proportional between them is 18. The numbers are a) b) c) d)
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10. | The sides of an equilateral triangle are shortened by 12 units 13 units and 14 units respectively and a right angle triangle is formed. The side of the equilateral triangle is a) b) c) d)
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11 | A distributor of apple Juice has 5000 bottle in the store that it wishes to distribute in a month. From experience it is known that demand D (in number of bottles) is given by The price per bottle that will result zero inventory is a) ₹ 3b) ₹ 5 c) ₹ 2d) none of these
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12. | The sum of two irrational numbers multiplied by the larger one is 70 and their difference is multiplied by the smaller one is 12; the two numbers are a) b) c) d) none of these |
2.1.11 SOLUTION OF CUBIC EQUATION
On trial basis putting if some value of x stratifies the equation then we get a factor. This is a trial and error method. With this factor to factorise the LHS and then other get values of x.
ILLUSTRATIONS:
1. | Solve Putting L.H.S is Zero. So is a factor of We write in such a way that becomes its factor. This can be achieved by writing the equation in the following form. or or or or or or or
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2. | Solve for real SOLUTION: By trial we find that makes the LHS zero. So is a factor of We write as or or Either Or i.e., as is not real, is the required solution. |
UNIT-I : EXERCISE (I)
Choose the most appropriate option (a), (b), (c) or (d)
1. The solution of the cubic equation is given by the triplet : a) b) c) d) |
2. The cubic equation has 3 roots namely. a) b) c) d)
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3. are the factors of the left–hand side of the equation a) b) c) d)
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4. The equation has got 3 roots and hence the factors of the left–hand side of the equation are a) b) c) d)
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5. The roots of the equation are a) b) c) d)
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6. The roots of are a) b) c) d)
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7. The satisfying value of are a) b) c) d)
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8. The roots of the cubic equation are a) b) c) d)
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9. If then value of is given by a) b) c) d) none of these\ |
10. The rational root of the equation is a) b) c) d) |
SUMMARY
- A simple equation in one unknown xis in the form
Where a, b are known constants and
- The general form of a linear equations in two unknowns x and y is wherea, b are non-zero coefficients and c is a constant. Two such equations and form a pair of simultaneous equations in x and y. A value for each unknown which satisfies simultaneously both the equations will give the roots of theequations.
- Elimination Method: In this method two given linear equations are reduced to a linear
equation in one unknown by eliminating one of the unknowns and then solving for the other unknown.
- Cross Multiplication Method:Let two equations be:
- An equation of the form where xis a variable and a, b, c are constantswith is called a quadratic equation or equation of the second degree.
When b=0 the equation is called a pure quadratic equation; when b = 0 the equation is
called an affected quadratic.
- The roots of a quadratic equation:
- The Sum and Product of the Roots of quadratic equation
- sum of roots
- product of the roots
- To construct a quadratic equation for the equation we have
- (Sum of the roots ) x + Product of the roots = 0
- Nature of the roots
- If the roots are real and equal;
- if then the roots are real and unequal (or distinct);
- If then the roots are imaginary;
- If is a perfect square the roots are real, rational and unequal (distinct);
- If but not a perfect square the rots are real, irrational and unequal.
Since discriminates the roots is called the discriminant in theequationas it actually discriminates between the roots.
ANSWERS
UNIT –I
Exercise (A)
1. | (b) | 2. | (a) | 3. | (c) | 4. | (c) | 5. | (b) | 6. | (d) | 7. | (a) | 8. | (d) |
9. | (c) |
Exercise (B)
1. | (c) | 2. | (b) | 3. | (a) | 4. | (b) | 5. | (c) | 6. | (a) | 7. | (d) | 8. | (d) |
9. | (a) | 10. | (c) | 11. | (c) | 12. | (a) |
Exercise (C)
1. | (b) | 2. | (c) | 3. | (a) | 4. | (a) | 5. | (d) | 6. | (a) | 7. | (b) | 8. | (c) |
9. | (b) | 10. | (d) |
Exercise (D)
1. | (a) | 2. | (c) | 3. | (a) | 4. | (d) | 5. | (a) | 6. | (c) | 7. | (a) | 8. | (c) |
9. | (b) | 10. | (d) |
Exercise (E)
1. | (b) | 2. | (a) | 3. | (d) | 4. | (c) | 5. | (b) | 6. | (c) | 7. | (a) | 8. | (a) |
9. | (c) | 10. | (b) | 11. | (a) |
Exercise (F)
1. | (d) | 2. | (d) | 3. | (b) | 4. | (b) | 5. | (a) | 6. | (c) | 7. | (d) | 8. | (b) |
9. | (a) | 10. | (b) | 11. | (d) | 12. | (c) | 13. | (d) |
Exercise (G)
1. | (b) | 2. | (d) | 3. | (a) | 4. | (d) | 5. | (b) | 6. | (b) | 7. | (a) | 8. | (c) |
9. | (c) | 10. | (a) |
Exercise (H)
1. | (c) | 2. | (b) | 3. | (a) | 4. | (b) | 5. | (c) | 6. | (d) | 7. | (a) | 8. | (b) |
9. | (c) | 10. | (a) | 11. | (a) | 12. | (c) |
Exercise (I)
1. | (c) | 2. | (b) | 3. | (b) | 4. | (b) | 5. | (b) | 6. | (a) | 7. | (d) | 8. | (b) |
9. | (a) | 10. | (c) |
ADDITIONAL QUESTION BANK
1.Solving equation x2-(a+b)x+ab=0 are,value(s) of x (a) a, b b) a c) b d) None
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2. Solving equation x2-24x+135=0 are, value(s) of x a) 9,6 b) 9,15 c) 15,6 d) None
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3. If (x/b)+(b/x)=(a/b)+(b/a) the roots of the equation are a) a,b2 /a b) a2 ,b/a2 c) a2 ,b2 /a d) a,b2
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4. Solving equationn we get roots as a) ±1 b) +1 c) -1 d) 0 |
5. Solving equation we get roots as
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6. Solving equation we get roots as a) b) c) d) None
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7. Solving equation following roots are obtained a) b) c) d) None
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8. Solving equation following roots are obtained a) b) c) d)
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9. Solving equation following roots are obtained a) b) c) d)
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10. Solving equation following roots are obtained a) b) c) all of above d) None
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11. Solving equation following roots are obtained a) b) both c) onlyd)
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12. Solving equation are, values of x a) b) c) d)
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13. Solving equation the roots available are a) b) c) d) None
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14. Solving equation the roots available are a) b) c) d) None
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15. Solving equation the value of z works out to a) b) c) d) |
16. Solving equation the following value of z are obtained a) b) c) d)
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17. When the value of z is given by a) b) c) d)
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18. Solving equation following roots are obtained a) b) c) d)
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19. Solving equation following roots are obtained a) b) c) d)
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20. Solving equation following roots are obtained a) b) c) d) None
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21. If the value of x is a) b) c) d)
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22. If then the value of x are a) b) c) d) None
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23. If then the value of x are a) b) c) d) |
24. If the value of is a) b) c) d)
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25. Solving equation we get roots as follows a) b) c) d) None |
26. Solving equation we get roots as follows a) b) c) d) |
27. Solving equation we get roots as under a) b) c) d)
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28. If are the roots of equation then the equation with roots and is a) b) c) d)
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29. If are the roots of equation then the equation with roots and is a) b) c) d) None
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30. If are the roots of equation then the equation with roots and is a) b) c) d) None
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31. The condition that one of the roots of is twice the other is a) b) c) d)
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32. The condition that one of the roots of is thrice the other is a) b) c) d)
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33. If the roots of are in the ratiothen the value of is a) b) c) d)
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34. Solving and we get values of x and y as a) b) c) d)
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35. Solving and we get the roots as under a) b) c) d)
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36. Solving and we get the roots as under a) b) c) d)
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37. Solving and we get the roots as under a) b) c) d)
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38. Solving and we get the roots as under a) b) c) d) None
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39. Solving and we get the following roots a) b) c) d) None
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40. Solving and we get the following roots a) b) c) d)
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41. Solving and we get the following roots a) b) c) d)
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42. Solving and following roots are obtained a) b) c) d) None
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43. Solving and following roots are obtained a) b) c) only d) only
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44. Solving get the following roots a) b) c) d)
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45. Solving get the following roots a) b) c) d) None
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46. It is being given that one of the roots is half the sum of the order two solving get the following roots : a) b) c) d) |
47. Solve given that the roots are in arithmetical progression a) b) c) d)
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48. Solve given that the roots are in geometrical progression a) ½, b) c) d)
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49. Solve given that the product of the two roots is 12 a) b) c) d)
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50. Solve given that two of its roots being in the ratio of 3:4 a) b) c) d)
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ANSWERS
1. | (a) | 18. | (a) | 35. | (a) |
2. | (b) | 19. | (b) | 36. | (a) |
3. | (a) | 20. | (a) | 37. | (b) |
4. | (b) | 21. | (d) | 38. | (c) |
5. | (b) | 22. | (d) | 39. | (a) |
6. | (c) | 23. | (d) | 40. | (a), (b) |
7. | (a) | 24. | (a) | 41. | (a) |
8. | (a) | 25. | (b) | 42. | (c) |
9. | (a) | 26. | (d) | 43. | (a) |
10. | (c) | 27. | (d) | 44. | (c) |
11. | (a) | 28. | (a) | 45. | (a) |
12. | (c) | 29. | (c) | 46. | (b) |
13. | (a) | 30. | (a) | 47. | (c) |
14. | (b) | 31. | (c) | 48. | (b) |
15. | (c) | 32. | (a) | 49. | (b) |
16. | (a) | 33. | (a) | 50. | (a) |
17. | (d) | 34. | (b) |
*This article contains all topics about Equations and matrices- Business Mathematics and Logical Reasoning & Statistics
For notes on all CA foundation topics, you can visit this article CA foundation note