Chapter 4 QUADRATIC EQUATIONS

1. Check whether the following are Quadratic Equations.

(i) = 2 (x − 3)

(ii) = (−2) (3 − x)

(iii) (x − 2) (x + 1) = (x − 1) (x + 3)

(iv) (x − 3) (2x + 1) = x (x + 5)

(v) (2x − 1) (x − 3) = (x + 5) (x − 1)

(vi)

(vii)

(viii) 

Explanation:

 (i) = 2 (x − 3)

⇒ = 2x – 6

⇒ = 0

Equation degree in this case is 2.


It is a quadratic equation as a result.

(ii) = (−2) (3 − x)

⇒ = −6 + 2x

⇒ − 2x + 6 = 0

⇒ = 0

Equation degree in this case is 2.


It is a quadratic equation as a result.

(iii) (x − 2) (x + 1) = (x − 1) (x + 3)

⇒ =

⇒

⇒ x − 2x – 2 − 3x + x + 3 = 0

⇒ −3x + 1 = 0

Equation degree in this case is 1.


It is a not quadratic equation as a result.

(iv) (x − 3) (2x + 1) = x (x + 5)

⇒ =

⇒

⇒ = 0

Equation degree in this case is 2.


It is a quadratic equation as a result.

(v) (2x − 1) (x − 3) = (x + 5) (x − 1)

⇒ =

⇒

⇒ x2 − 11x + 8 = 0

Equation degree in this case is 2.


It is a quadratic equation as a result.

(vi)

⇒ =

⇒

⇒ 7x – 3 = 0

Equation degree in this case is 1.


It is not a quadratic equation as a result.

(vii)

⇒ =

⇒ =

⇒

⇒

Equation degree in this case is 3.


It is not a quadratic equation as a result.

(viii)

⇒ = (x) (2) (x − 2)

⇒ =

⇒ = 0

The equation degree in this case is 2.

It is a quadratic equation as a result.

2. Represent the following situations in the form of Quadratic Equations:

(i) The area of the rectangular plot is. The length of the plot (in meters) is one more than twice its breadth. We need to find the length and breadth of the plot.

(ii) The product of two consecutive numbers is 306. We need to find the integers.

(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) after 3 years will be 360. We would like to find Rohan’s present age.

(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Explanation:


(i) We are informed what a rectangle plot's area is.

Let the width of the rectangle be x metres.

The length is one and a half times the width.

Hence, the rectangular plot's length is (2x + 1) metres.

The area of a rectangle equals length times breadth

⇒ 528 = x (2x + 1)

⇒ 528 =

⇒– 528 = 0

This is a Quadratic Equation.

(ii) Let x and (x + 1) be two numbers that follow each other.

The answer is given as x (x + 1) = 306.

⇒ = 306

⇒– 306 = 0

This is a Quadratic Equation.

(iii) Rohan's age right now is x years.

Let the mother of Rohan's current age be (x + 26) years.

Rohan's age after three years equals (x + 3) years.

After three years, the mother of Rohan is x + 26 + 3 = (x + 29) years old.

Considering the above circumstance:

(x + 3) (x + 29) = 360

⇒ = 360

⇒ = 0

This is a Quadratic Equation.

(iv) Let the train's speed be x km/h.

480 kilometres travelled by train in 480 hours.

The amount of time would have been (480 x 8) hours if the speed had been 8 km/h less.

Assuming the conditions are met, the duration would have been reduced by 3 hours if the speed had been 8 km/h lower.

Therefore, 480x – 8 = 480x + 3

⇒ 480 (1x – 8 − 1x) = 3

⇒ 480 (x – x + 8) (x) (x − 8) = 3

⇒ 480 × 8 = 3 (x) (x − 8)

⇒ 3840 =

⇒ = 0

By dividing the equation by 3, we get

⇒ = 0

This is a Quadratic Equation.

3. Find the roots of the following Quadratic Equations by factorization.

(i)

(ii)

(iii)

(iv)

(v)

Explanation:


 (i)x2 − 3x – 10 = 0

⇒

⇒x (x − 5) + 2 (x − 5) = 0

⇒(x − 5) (x + 2) = 0

⇒x = 5, −2

(ii)

⇒

⇒2x (x + 2) – 3 (x + 2) = 0

⇒(2x − 3) (x + 2) = 0

⇒x =

(iii)

⇒

⇒

⇒

⇒

⇒

⇒

(iv) = 0

⇒

⇒

⇒

⇒4x (4x − 1) – 1 (4x − 1) = 0

⇒(4x − 1) (4x − 1) = 0

⇒

(v)

⇒

⇒10x (10x − 1) – 1 (10x − 1) = 0

⇒(10x − 1) (10x − 1) = 0

⇒x = 

4. Solve the following situations mathematically:

(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.

(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ₹750. We would like to find out the number of toys produced on that day.

Explanation:


(i) Let's say that John has x marbles in total.

Jivanti's marble count is therefore 45 - x.

After losing 5 marbles apiece, the Number of marbles John has = x - 5

Jivanti's marble count is 45 - x - 5 = 40 - x.

Assuming that their marbles have a product of 124.

∴ (x – 5)(40 – x) = 124

⇒ x2 – 45x + 324 = 0

⇒ x2 – 36x – 9x + 324 = 0

⇒ x(x – 36) -9(x – 36) = 0

⇒ (x – 36)(x – 9) = 0

Thus, we can say,

x – 36 = 0 or x – 9 = 0

⇒ x = 36 or x = 9

Therefore,

If 36 John's marbles,

Jivanti's marbles then equal 45 - 36 = 9.

And if John's marbles equal nine

Jivanti's marbles then equal 45 - 9 = 36.

(ii) Assume that x toys are made every day.

Hence, each toy's manufacture cost is Rs (55 – x)

Assuming that the toys' entire manufacture cost was Rs. 750

∴ x(55 – x) = 750

⇒ x2 – 55x + 750 = 0

⇒ x2 – 25x – 30x + 750 = 0

⇒ x(x – 25) -30(x – 25) = 0

⇒ (x – 25)(x – 30) = 0

Thus, either x -25 = 0 or x – 30 = 0

⇒ x = 25 or x = 30

As a result, either 25 or 30 toys will be created each day.

5. Find two numbers whose sum is 27 and whose product is 182.

Explanation:


Let x be the first number and (x + 27) be the second.

The sum of two numbers, under the conditions stated, is 182.

Therefore,

x (27 − x) = 182

⇒

⇒

⇒

⇒x (x − 14) – 13 (x − 14) = 0

⇒(x − 14) (x − 13) = 0

⇒x = 14, 13

The first number, therefore, equals 14 or 13.

Therefore the second number is equal to = 27 - x = 27 - 14 = 13 or = 27 - 13 = 14.

13 and 14 are two numbers as a result.

6. Find two consecutive positive integers, the sum of whose squares is 365.

Explanation:


Let x be the first number and (x + 1) be the second.

Considering the situation,

⇒

⇒

Dividing equation by 2

⇒

⇒

⇒x (x + 14) – 13 (x + 14) = 0

⇒ (x + 14) (x − 13) = 0

⇒x = 13, −14

As a result, the first number is 13 (we ignore the negative number -14).

Number two: x + 1 = 13 + 1 = 14.

Hence, the total of the squares of two successive positive integers 13 and 14 is 365.

7. The altitude of a right triangle is 7 cm less than its base. If, hypotenuse is 13 cm. Find the other two sides.

Explanation:


Let the triangle's base be x cm and its height be (x 7) cm.

Given that the triangle's hypotenuse measures 13 cm

The Pythagorean Theorem states that

=

⇒

⇒169 =

⇒= 0

Dividing equation by 2

⇒

⇒

⇒x (x − 12) + 5 (x − 12) = 0

⇒(x − 12) (x + 5)

⇒x = −5, 12

We rule out x = 5 since a triangle's side length cannot be negative.

Hence, the triangle's base is 12 cm.

Triangle's altitude is equal to (x-7) = 12 - 7 = 5 cm.

8. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If, the total cost of production on that day was Rs. 90, find the number of articles produced and the cost of each article.

Explanation:


Let's say that each article's production costs are Rs x.

The whole cost of production for that day is presented to us as Rs 90.

Therefore, 90/x is the number of articles produced overall that day.

Considering the circumstances,

⇒

⇒

⇒

⇒

⇒

⇒x (x − 15) + 12 (x − 15) = 0

⇒(x − 15) (x + 12) = 0⇒x = 15, −12

Cost cannot be negative, thus we rule out x = 12.

As a result, the cost of producing each article, x, is Rs 15.

quantity of articles produced on a given day = 6

9. Find the roots of the following quadratic equations if they exist by the method of completing the square.

(i)

(ii)

(iii)

(iv) 

Explanation:

(i)

To set the coefficient of x2 equal to 1, we first divide the equation by 2.

We obtain this by dividing the middle term of the equation by 2x.

From the equation, we add and subtract the square of   


⇒

⇒

⇒

Using the Square Root formula on both sides,

⇒

⇒

Therefore,

(ii)

Dividing the equation by 2,

Using the square-filling approach,

⇒

⇒

⇒

square roots on each side,

⇒

⇒

Therefore,

(iii)

Dividing the equation by 4,

Using the square-filling approach,

⇒

⇒

⇒

square roots on each side,

⇒

⇒

(iv)

Dividing the equation by 2,

Using the square-filling approach,

 ⇒

⇒

⇒

⇒

square roots on each side

Because there is no square root of a negative number, the right side does not exist.

As a result, the quadratic equation has no solution.

10. Find the roots of the following Quadratic Equations by applying the quadratic formula.

(i)

(ii)

(iii)

(iv) 

Explanation:


(i)

Comparing quadratic equation with general form , we get a = 2, b = -7 and c = 3

Putting these values in the quadratic formula

⇒

⇒

⇒

⇒ x = 3, ½

(ii)

Comparing quadratic equation with the general form , we get a = 2, b = 1 and c = −4

Putting these values in the quadratic formula

⇒

⇒

(iii)

Comparing quadratic equation with the general form, , we get a = 4, b = and c = 3

Putting these values in the quadratic formula

⇒

⇒

A quadratic equation has two roots. Here, both the roots are equal.

Therefore,

(iv)

Comparing quadratic equation with the general form , we get a = 2, b = 1 and c = 4

Putting these values in the quadratic formula

⇒

But, the square root of a negative number is not defined.

Therefore, Quadratic Equation has no solution.

11. Find the roots of the following equations:

(i)

(ii) 

Explanation:

(i)

⇒

⇒

⇒

Comparing the equation with the general form,

We get a = 1, b = −3 and c = −1

Using quadratic formulas to solve equations,

⇒

⇒

(ii)

⇒

⇒

⇒

⇒

Comparing the equation with the general form,

We get a = 1, b = −3 and c = 2

Using quadratic formulas to solve equations,

⇒

⇒

⇒ x = 2, 1

12. The sum of the reciprocals of Rehman’s age (in years) 3 years ago and 5 years from now is 1/3. Find his present age.

Explanation:


Let Rehman's age at this time be x years.

The age of Rehman three years ago equals (x -3) years.

Rehman's age after five years equals (x + 5) years.

In light of the stated circumstance:

⇒

⇒ 3 (2x + 2) = (x − 3) (x + 5)

⇒ 6x + 6 =

⇒

⇒

Comparing quadratic equation 0 with the general form,

We get a = 1, b = −4 and c = −21

Using quadratic formula

⇒

⇒

⇒

⇒ x = 7, −3

x=-3 is discarded. Age cannot be negative, therefore.

Rehman is therefore currently 7 years old.


13. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.

Explanation:


Let Shefali's math grades equal x.

Let the English grade for Shefali be 30− x.

Her grades would have been = x + 2 if she had received 2 additional marks in mathematics.

If she had received 3 fewer points in English, her English score would have been = 30 - x- 3 = 27- x.

Considering the above circumstance:

(x + 2) (27 − x) = 210

⇒ = 210

⇒

Comparing quadratic equations with the general form,

We get a = 1, b = −25 and c = 156

Applying Quadratic Formula

⇒

⇒

⇒

⇒ x = 13, 12

Shefali's math grades are therefore 13 or 12.

Shefali received a grade of 30 - x = 30 - 13 = 17.

Instead, Shefali's English marks are 30 - x = 30 - 12 = 18.

Her scores in math and English are thus (13, 17) or (12, 18).

14. The diagonal of a rectangular field is 60 meters more than the shorter side. If, the longer side is 30 meters more than the shorter side, find the sides of the field.

Explanation:


Let the rectangle's shorter side equal x metres.

Let the rectangle's diagonal equal (x + 60) metres.

Let the rectangle's longer side equal (x + 30) metres.

The Pythagorean Theorem states that

=

⇒ =

⇒

Comparing equation with standard form,

We get a = 1, b = −60 and c = −2700

Applying quadratic formula

⇒

⇒

⇒

⇒ x = 90, –30

We disregard -30. Considering that length cannot be negative.

Hence, x = 90, meaning that the shorter side's length is 90 metres.

And the longer side is x + 30 metres, which is 90 metres plus 30 metres.

As a result, the sides are 90 and 120 metres long.

15. The difference between the squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Explanation:


Let x be the smaller number and y be the larger number.

In light of the situation:

We are also informed that the square of a smaller number is eight times that of a larger number.

⇒ … (2)

Equation (2) is obtained by applying it to (1).

⇒

Comparing the equation with the general form,

We get a = 1, b = −8 and c = −180

Using quadratic formula

⇒

⇒

⇒

⇒ y = 18, −10

In order to find a lesser number, use equation (2):

⇒ = 144

⇒ x = ±12

And, = {No real solution for x}

Therefore, the two numbers are either (12, 18) or (-12, 18).

16. A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Explanation:


Let the train's speed be x km/hr.

The train would have arrived in half the time if its speed had been 5 km/hr higher.

Thus, given this circumstance

⇒

⇒

⇒

⇒

Comparing equation with the general equation,

We get a = 1, b = 5 and c = −1800

Applying quadratic formula

⇒

⇒

⇒

⇒ x = 40, −45

since a train's speed cannot be negative. As a result, we ignore x = -45.

Hence, the train's speed is 40 km/hr.

17. Two water taps together can fill a tank in hours. The tap of a larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time at which each tap can separately fill the tank.

Explanation:


Assume that it takes x hours to fill the tank using a tap with a smaller diameter.

Let the amount of time needed to fill the tank with a larger-diameter tap be (x - 10) hours.

This indicates that a smaller-diameter tap may fill a portion of a tank in an hour. … (1)

And a larger diameter tap can fill a portion of the tank in an hour. (2)

Tanks may be filled in 758 hours when two taps are used together.

In 1 hour, they fill part of the tank, … (3)

From (1), (2), and (3),

⇒

⇒ 75 (2x − 10) =

⇒ 150x – 750 =

⇒

⇒

Comparing the equation with the general equation,

We get a = 4, b = −115 and c = 375

Applying quadratic formula

⇒

⇒

⇒

⇒

⇒ x = 25, 3.75

The time required by the larger tap is equal to x - 10 = 3.75 - 10 = 6.25 hours.

Time cannot go backward. As a result, we disregard this value.

The time required by the larger tap is equal to x - 10 = 25 - 10 = 15 hours.

As a result, the larger tap requires 15 hours and the smaller tap requires 25 hours.

18. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If, the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of two trains.

Explanation:


Let the average passenger train speed be x km/h.

Let the express train's average speed be (x + 11) km/h


hours are needed for a passenger train to go 132 kilometers.


hours are needed for an express train to go 132 kilometers.


Considering the stated circumstance,

⇒

⇒

⇒ 132 (11) = x (x + 11)

⇒

⇒

Comparing equation with general quadratic equation , we get a = 1, b = 11 and c = −1452

Applying Quadratic Formula

⇒

⇒

⇒

⇒

⇒ x = 33, −44

Given that speed cannot be negative. Hence, the passenger train's speed is 33 km/h.

And the express train's speed is equal to x + 11 = (33 + 11)=44 km/h.

19. Sum of the areas of two squares is 468 m2. If, the difference between their perimeters is 24 meters, find the sides of the two squares.

Explanation:


Let the first square's perimeter be x metres.

Let the second square's perimeter equal (x + 24) metres.

The first square's initial side length is measured in metres. {Perimeter of square = }

m for the second square's side length

The first square's area equals its side-by-side.

Size of the second square:

Considering the above circumstance:

⇒

⇒

⇒ =

⇒ = 7488

⇒ = 0

⇒ = 0

Comparing equation =0 with standard form,

We get a = 1, b =24 and c = −3456

Applying Quadratic Formula

⇒

⇒

⇒

⇒ x = 48, −72

A square's perimeter cannot be negative. As a result, we ignore x=-72.

Hence, the perimeter of the first square = 48 metres

Moreover, the second square's perimeter is equal to x + 24= (48 + 24) = 72 metres.

the first square's side =

The side of second square also equals = 

20. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them.

(i)

(ii)

(iii) 

Explanation:


(i)

Comparing this equation with general equation ,

We get a = 2, b = −3 and c = 5

Discriminant = (2) (5)

= 9 – 40 = −31

Discriminant is less than 0 which means equation has no real roots.

(ii)

Comparing this equation with general equation ,

We get a = 3, b = and c = 4

Discriminant = = − 4 (3) (4)

= 48 – 48 = 0

Discriminant is equal to zero which means equations has equal real roots.

Applying quadratic to find roots,

Because, equation has two equal roots, it means

(iii)

Comparing equation with general equation ,

We get a = 2, b = −6, and c = 3

Discriminant = (2) (3)

= 36 – 24 = 12

Value of discriminant is greater than zero.

Therefore, equation has distinct and real roots.

Applying quadratic formula to find roots,

⇒

⇒ 

21. Find the value of k for each of the following quadratic equations, so that they have two equal roots.

(i)

(ii) kx (x − 2) + 6 = 0

Explanation:


(i)

Only when the value of the discriminant is equal to zero do we know that a quadratic equation has two equal roots.

Comparing equation with general quadratic equation , we get a = 2, b = k and c = 3

Discriminant = (2) (3) =

Putting discriminant equal to zero

⇒

⇒

(ii)kx (x − 2) + 6 = 0

⇒

Comparing quadratic equation with general form , we get a = k, b = −2k and c = 6

Discriminant = (k) (6) =

Only when the discriminant equals zero do we know that the two roots of a quadratic equation are equivalent.

Putting discriminant equal to zero

⇒ 4k (k − 6) = 0⇒ k = 0, 6

According to the basic definition, a quadratic equation is an equation of the form where a ≠ 0.

Thus, we cannot have k = 0 in equation.

As a result, we ignore k = 0.

Hence, the solution is k = 6.

22. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is. If so, find its length and breadth.

Explanation:


Assume that the rectangular mango grove's width is x metres.

Let the rectangular mango grove's length equal 2x metres.

Rectangle area = =

Considering the above circumstance:

⇒ = 0 = 0

Comparing equation with general form of quadratic equation , we get a = 1, b = 0 and c = −400

Discriminant (1) (−400) = 1600

Equation with two distinct real roots has discriminant bigger than 0, which is greater than 0.

Consequently, a rectangular grove might be created.

Applying quadratic formula, to solve equation,

⇒ x = 20, −20

Because the width of a rectangle cannot be negative, we ignore negative values of x.

Hence, x = rectangle's width, which is 20 metres

rectangle length = 2x = = 40 metres.

23. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.

Explanation:


Let the ages of the first and second friends be x and (20-x) respectively.

Age of first friend four years ago equals (x-4) years.

Age of second friend four years ago: (20 − x) − 4 = (16 − x) years

Considering the situation,

(x − 4) (16 − x) = 48

⇒ = 48

⇒

⇒

Comparing equation, with general quadratic equation , we get a = 1, b = −20 and c = 112

Discriminant = (1) (112) = 400 – 448 = −48 < 0

Since discriminant is smaller than zero, this equation has no true roots.

The give situation is therefore impossible.

24. Is it possible to design a rectangular park of perimeter 80 metres and area 400 m2. If so, find its length and breadth.

Explanation:


Let length of park = x metres

We are given area of rectangular park =

Therefore, breadth of park = metres

Hence, the area of a rectangle is equal to the sum of its length and breadth.Park's perimeter is equal to 2 (length + breath) metres = metres

When provided, 80 metres is the rectangle's perimeter.

In light of the situation:

⇒

⇒

⇒

⇒

Comparing equation, with general quadratic equation , we get a = 1, b = −40 and c = 400

Discriminant = (1) (400) = 1600 – 1600 = 0

Discriminant is equal to 0.

As a result, the equation's two real and equal roots allow for the creation of a rectangular park with an 80-meter perimeter and an area .

Using quadratic formula to solve equation,

In this case, both roots add up to 20.

Hence, the rectangular park's length is 20 metres.

Breadth of rectangular park =

Chapter 4 QUADRATIC EQUATIONS