QUADRATIC EQUATIONS (LECTURE 3)

13TH MAY
LESSON 3 QUADRATIC EQNS.

GOOD MORNING!!
IMPORTANT e CIRCULAR
*We at St Columba's School, are ready to take the next step - and move towards using Google Classroom as a tool to enhance the teaching learning environment.

*This shift involves the school giving each student a unique email address, which he will need to setup ( the teachers will assist him in doing the same). This email will facilitate each student to interact with his teachers.

* The student should use this email ID only for:
-interacting with  teachers on school /subject related matters
-keep the words and expressions as he would- if he  were interacting with us as a Columban student
-use the facilities connected to this email ID only for google meets organised with the permission of a teacher 
-he should have a copy of the permission given ( when and by which teacher and for what purpose)
-do not access any of the other facilities connected with this email address unless he first seeks the permission of a teacher
(He should have a copy of the permission given - when and by which teacher and for what purpose).

*Please note that Google and St Columba's have legal obligations by giving you access to this email address and hence,
you will need to sign an agreement when creating this email ID. 

*Be informed that the history of use of this ID shall have a digital footprint !



IN THE PREVIOUS CLASS YOU LEARNT 
how to solve quadratic equations by factorization method 



Today 'S learning outcomes are:
I will be able to:

comprehend and translate the word problems as quadratic equations and solve them  using factorization method.


TODAY WE WILL COVER Q 4,5 AND 6 OF EX 4.2

YOU ALL WILL VIEW THE VIDEO SOLUTIONS OF THESE THREE QUESTIONS  AND TRY TO COMPREHEND THE SOLUTION.

                                                VIEW AND UNDERSTAND 


1. CLICK ON THE LINK  Q4

2 CLICK ON THE LINK Q5

3.CLICK ON THE LINK Q6

NOW, IT'S TIME FOR SOME WRITTEN WORK!!

MEETING ID  8:00am to 9:00am  


                                                         CLASS WORK
                                      
   

                                                                   EX 4.2



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

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

According to given condition,



⇒
⇒
Dividing equation by 2
⇒
⇒
⇒x (x + 14) – 13 (x + 14) = 0
⇒ (x + 14) (x − 13) = 0
⇒x = 13, −14
Therefore, first number = 13 {We discard -14 because it is negative number)
Second number = x + 1 = 13 + 1 = 14
Therefore, two consecutive positive integers are 13 and 14 whose sum of squares is equal to 365.

5.The altitude of right triangle is 7 cm less than its base. If, hypotenuse is 13 cm. Find the other two sides.
Ans. Let base of triangle be x cm and let altitude of triangle be (x − 7) cm
It is given that hypotenuse of triangle is 13 cm
According to Pythagoras Theorem,
= 
⇒
⇒169 =
⇒= 0
Dividing equation by 2
⇒
⇒
⇒x (x − 12) + 5 (x − 12) = 0
⇒(x − 12) (x + 5)
⇒x = −5, 12
We discard x = −5 because length of side of triangle cannot be negative.
Therefore, base of triangle = 12 cm
Altitude of triangle = (x − 7) = 12 – 7 = 5 cm

6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that 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.
Ans. Let number of articles be' n'
  
 so,cost of production of each article will be  2n +3
We are given total cost of production on that particular day = Rs 90

According to the given conditions,
⇒ n(2n +3)  =90
⇒ 2n²  +3n = 90
⇒ 2n²  +3n -90 = 0
⇒ 2n² -12n  +  15 n  -90 = 0
⇒ 2n( n-6)  +15(n -6) =0

⇒  2 n +15 =0 , n -6 =0
⇒  n = -15/2 , n =6
no. of articles  cannot be in negative, therefore, we discard n  = −15/2
Number of articles produced on that particular day = 6




Look at the solved eg 6
 A  charity trust decides to build a prayer hall having a carpet area of 300 square metres with its length one metre more than twice its breadth. What should be the length and breadth of the hall? 


What is given?

Area of prayer hall  = 300 m²

What is unknown?

length and breadth of the hall

Working?

What should be assumed as 'x'

The answer is breadth

so, let the breadth be 'x'

and therefore length will be  2x +1

ESTABLISH A RELATION BETWEEN UNKNOWNS AND GIVEN INFORMATION

WE , KNOW

Area of rectangle  = length  x breadth

SUBSTITUTE THE VALUES

300  = x  (2x +1)  ....(dont forget the brackets)

simplifying the bracket

==>300  =  2x ² +x 

WRITE IN THE STANDARD FORM

=> 
2x ² +x - 300  =0

 Applying the factorisation method 

=>2x² – 24x + 25x – 300 = 0

=> 2x (x – 12) + 25 (x – 12) = 0

i.e., (x – 12)(2x + 25) = 0

So, the roots of the given equation are x = 12 or x = – 12.5. 

Since x is the breadth of the hall, it cannot be negative.

Thus, the breadth of the hall is 12 m. Its length = 2x + 1 = 25 m.


    





HOME WORK


Q1  Had Kavita scored 10 more marks in her Mathematics test out of 30 marks, 9 times these marks would have been the square of her actual marks. How many marks did she get in the test?
   Q2 A natural number when subtracted from 28, becomes equal to 160 times its reciprocal. Find the number.

Q3    Find two consecutive odd positive integers, sum of whose squares is 290.





WE END  TODAY'S CLASS HERE!!
TAKE CARE AND KEEP SAFE!!









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