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)
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.
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)
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.
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)
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)
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)
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)
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
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 =
Also Read:Quadratic Equation Class 10 Extra Questions
Chapter 4 QUADRATIC EQUATIONS
FAQs
What is a quadratic equation for 10th class? ›
A quadratic equation in the variable x is an equation of the form ax2 + bx + c= 0, where a, b, c are real numbers, a 0. For example, 2x2 + x – 300 = 0 is a quadratic equation. Similarly, 2x2 – 3x + 1 = 0, 4x – 3x2 + 2 = 0 and 1 – x2 + 300 = 0 are also quadratic equations.
What are the 4 quadratic solutions? ›The four methods of solving a quadratic equation are factoring, using the square roots, completing the square and the quadratic formula.
How many exercises are there in Chapter 4 of Class 10 maths? ›How many exercises are there in chapter 4 of Class 10th Maths? There are in all 4 exercises in class 10 mathematics chapter 4 (Quadratic equations). In first exercise (Ex 4.1), there are only 2 questions (Q1 having 8 parts and Q2 having 4 parts). In second exercise (Ex 4.2), there are in all 6 questions.
Which chapter is quadratic equation? ›NCERT Solutions for class 10 maths chapter 4 of the section, concentrates on the essential concepts such as the quadratic equation definition, standard form of a quadratic equation, nature of roots, concept of discriminants, quadratic formula, factorization technique of solving a quadratic equation, and completing the ...
What is the real life example of quadratic equation Class 10? ›Quadratic equations are used in various real-life situations such as calculating profit or the speed of an object. Real-life examples of quadratic functions are throwing a ball, profit and loss of product, and so on.
What is quadratic equation in math Grade 9? ›It is a polynomial equation of degree two that can be written in the form ax2 + bx + c = 0, where a, b, and c are real numbers and a ≠ 0.
What are the 5 different ways to solve a quadratic equation? ›- FACTORING. Set the equation equal to zero. ...
- PRINCIPLE OF SQUARE ROOTS. If the quadratic equation involves a SQUARE and a CONSTANT (no first degree term), position the square on one side and the constant on the other side.
- COMPLETING THE SQUARE. ...
- QUADRATIC FORMULA.
There are three basic methods for solving quadratic equations: factoring, using the quadratic formula, and completing the square.
What is the hardest chapter in maths? ›A chapter that most of the students find difficult in maths is trigonometry.
Which is the most easy chapter in maths class 10? ›CBSE Class 10 Math polynomial chapters involve the concepts like finding the degree of a polynomial, zeros of the polynomial, which are some of the easiest chapters in the syllabus. This chapter covers topics on solving the linear equation using different methods like elimination and substitution.
Which is the easiest chapter in class 10 maths? ›
Geometry: Geometry is also the easy part and holds the 19% weightage in CBSE class 10 maths exam i.e. 15 marks out of 80.
What grade is quadratic equations? ›9th-11th Grade Math - Quadratic Functions.
What are the 3 parts of a quadratic equation? ›The term ax2 is called the quadratic term (hence the name given to the function), the term bx is called the linear term, and the term c is called the constant term.
Is quadratic equations part of algebra? ›Quadratic equation is a part of algebra and here's a look at quadratic formula, geometry, quadratic factorization, and applications. In this equation, x represents a variable whereas a, b, and c are constants with a not equal to 0.
What are 5 quadratic equations examples? ›Examples of quadratic equations are: 6x² + 11x – 35 = 0, 2x² – 4x – 2 = 0, 2x² – 64 = 0, x² – 16 = 0, x² – 7x = 0, 2x² + 8x = 0 etc. From these examples, you can note that, some quadratic equations lack the term “c” and “bx.”
What are the roots of quadratic equation? ›The roots of a quadratic equation are the values of the variable that satisfy the equation. They are also known as the "solutions" or "zeros" of the quadratic equation. For example, the roots of the quadratic equation x2 - 7x + 10 = 0 are x = 2 and x = 5 because they satisfy the equation.
Why is quadratic equation important? ›Quadratic equation material is important because it makes it possible to establish relationships between various other mathematical material such as linear equations, functions, and polynomials (Sağlam & Alacacı, 2012).
What is the quadratic formula in high school math? ›The quadratic formula helps us solve any quadratic equation. First, we bring the equation to the form ax²+bx+c=0, where a, b, and c are coefficients. Then, we plug these coefficients in the formula: (-b±√(b²-4ac))/(2a) . See examples of using the formula to solve a variety of equations.
What does B mean in quadratic equation? ›B-value: The b-value is the middle number, which is the number next to and multiplied by the x; a change in the value of b affects the parabola and the resulting graph.
What are the 2 methods in solving quadratic equations? ›- Factorization.
- Completing the square method.
- Quadratic Equation Formula.
Which method is better to solve quadratic equation? ›
Quadratic formula – is the method that is used most often for solving a quadratic equation. If you are using factoring or the quadratic formula, make sure that the equation is in standard form.
What is the most method to solve the quadratic equation? ›Completing the square is a method that may be used for any quadratic equation. By adjusting your constant (c), you can create a perfect square on the left side of the equation. A perfect square can be factored into two identical binomials, which you can use to solve for any valid values of x.
What are the 2 types of quadratic formula? ›Standard form: The standard form of a quadratic equation is represented by y = a x 2 + b x + c where and are just the numbers. Factored form: The factored form of a quadratic equation is represented by y = ( a x + c ) ( b x + d ) where a , b , c , and are just the numbers.
Which is the easiest math chapter? ›One of the easiest CBSE Class 12 Maths chapters is Algebra which consists of Matrices and Determinants. You can quickly prepare this chapter to score good marks.
Why is math the hardest subject? ›Because math involves using plenty of multi-step processes to solve problems, being able to master it takes a lot more practice than other subjects. Having to repeat a process over and over again can quickly bore some children and this may make them become impatient with math.
What is the hardest math problem in the world with an answer? ›x3+y3+z3=k, with k being all the numbers from one to 100, is a Diophantine equation that's sometimes known as "summing of three cubes."
Which is the toughest chapter in 10th standard? ›4. What is the most difficult chapter in Class 10 Science? The toughest chapter in Class 10 Science is Heredity and Evolution.
Which is the shortest chapter in class 10 science? ›The simplest science chapter in class 10 is chemical reactions.
Which is the hardest subject in class 10? ›Students find Chemistry the easiest and Physics the toughest. Check paper analysis here. CBSE Board Exams 2023: Central Board of Secondary Education (CBSE) is conducting the annual exams for classes 10th and 12th.
Which subject is easiest in class 10 CBSE? ›English is one of the easiest subjects to learn & score high marks. This will also improve your overall percentage in the final result. Also, refer to the syllabus of the CBSE class 10 English. Below are some points that will help students on how to prepare for class 10 board exam CBSE 2023.
Which is the most scoring chapter in class 10 maths? ›
| Unit/Topic | Unit/Topic Name | Weightage |
|---|---|---|
| 1. | Polynomials | 4 Marks |
| 2. | Pair of Linear Equations in Two Variables | 3 Marks |
| 3. | Quadratic Equations | 9 Marks |
| 4. | Arithmetic Progressions | 4 Marks |
Circle, Parabola and Permutation and Combination are tough chapters of Class 11 Maths. Sequence and Series is another tough topic that needs more attention and preparation. Additionally, you must also study Coordinate Geometry and Integral Calculus for JEE Mains 2022.
What grade is algebra1? ›Some schools may offer Algebra I in either 9th/10th grade OR 11th/12th grade, but not both. Nonetheless, it is important that students have access to Algebra I sometime in their high school career.
Is the quadratic formula taught in algebra 1 or 2? ›Note that in Algebra I the focus is on linear, simple exponential, and quadratic equations.
What year level is quadratic equations? ›Year 10 Algebra Worksheet – Solving Quadratic Equations.
What does C stand for in quadratic equation? ›c is the constant term. The coefficient of the quadratic term, a, determines how wide or narrow the graphs are, and whether the graph turns upward or downward. Important Tidbit.
Is C the Y intercept? ›The equation y = mx + c is the general equation of any straight line where m is the gradient of the line (how steep the line is) and c is the y -intercept (the point in which the line crosses the y -axis).
When delta is 0? ›If Δ is equal to zero, the polynomial has only one real root. If Δ is less than zero, the polynomial has no real roots, only two distinct complex roots. A zero is the x value whereat the function crosses the x-axis. That is, it is the x-coordinate at which the function's value equals zero.
Which equation is not quadratic? ›(iv)x+5=3⇒0x2+x+2=0 which is not a quadratic equation because the coefficient of x2 (a)=0.
Is quadratic algebra 1? ›Quadratic equations (Algebra 1) – Mathplanet.
What is quadratic formula in high school? ›
The quadratic formula helps us solve any quadratic equation. First, we bring the equation to the form ax²+bx+c=0, where a, b, and c are coefficients. Then, we plug these coefficients in the formula: (-b±√(b²-4ac))/(2a) . See examples of using the formula to solve a variety of equations.
What is quadratic equation in math? ›A quadratic equation is a second-order polynomial equation in a single variable x ax2+bx+c=0. with a ≠ 0 . Because it is a second-order polynomial equation, the fundamental theorem of algebra guarantees that it has at least one solution. The solution may be real or complex.
Who gave quadratic formula class 10? ›Muhammad ibn Musa al-Khwarizmi was an Arab mathematician who wrote a revolutionary book on resolving quadratic equations. Q.
How do you solve quadratic equations in Class 10 word problems? ›- Denote the unknown quantities.
- Use the unknown quantities to Establish the problem of condition.
- Use the equations to establish a quadratic equation.
- Solve this equation to obtain the value of the unknown in which it belongs.
Read below for an explanation of the three main forms of quadratics (standard form, factored form, and vertex form), examples of each form, as well as strategies for converting between the various quadratic forms.
What grade is quadratic equations taught? ›9th-11th Grade Math - Quadratic Functions.
What are the 3 forms of quadratic functions? ›Forms of Quadratic Functions: Standard, Vertex & Factored.
What is C in a quadratic equation? ›c is the constant term. The coefficient of the quadratic term, a, determines how wide or narrow the graphs are, and whether the graph turns upward or downward. Important Tidbit. A positive quadratic coefficient causes the ends of the parabola to point upward.
What does B equal in math? ›B-value: The b-value is the middle number, which is the number next to and multiplied by the x; a change in the value of b affects the parabola and the resulting graph.
Who invented algebra? ›Muhammad ibn Musa al-Khwarizmi was a 9th-century Muslim mathematician and astronomer. He is known as the “father of algebra”, a word derived from the title of his book, Kitab al-Jabr. His pioneering work offered practical answers for land distribution, rules on inheritance and distributing salaries.
What is the famous quadratic formula? ›
f(x) = ax2 + bx + c, where a and b are coefficients, c is a constant value (also y-intercept (0, c)). Things to keep in mind when using the quadratic formula: The equation must be in standard form and must equal to 0 on one side.
How do you write a quadratic equation example? ›A quadratic function is a polynomial function of degree 2. So, y = x^2 is a quadratic equation, as is y = 3x^2 + x + 1. All of these are polynomial functions of degree 2.
How do you solve a quadratic system? ›- Isolate one of the two variables in one of the equations. ...
- Substitute the expression that is equal to the isolated variable from Step 1 into the other equation. ...
- Solve the resulting quadratic equation to find the x-value(s) of the solution(s).