QUADRATIC EQUATIONS (LECTURE 8)

  LESSON 8
QUADRATIC EQNS.
GOOD MORNING!!

IN THE PREVIOUS CLASS YOU LEARNT 

how to solve quadratic equations by Quadratic formula 

Today 'S learning outcomes are:

I will be able to:

1.comprehend and translate the word problems as quadratic equations and solve them 

2. recall about nature of roots of  quadratic equations.

TODAY WE WILL COVER  WORD PROBLEMS RELATED  TO SPEED/TIME;PERIMETER AND AREA; AND TIME/WORK

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

          VIEW AND UNDERSTAND

CLICK ON THE LINK TO VIEW Q8

CLICK ON THE LINK TO VIEW Q9

CLICK ON THE LINK TO VIEW Q10

CLICK ON THE LINK TO VIEW Q11

I HOPE YOU ALL HAVE GONE THRU THESE LINKS!

NOW IT'S TIME FOR SOME WRITTEN WORK

                EX4.3

SPEED/TIME

8. 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.

Ans. Let the speed of the train = x km/hr

If, speed had been 5 km/hr more, train would have taken 1 hour less.

So, according to this condition

 

⇒ 

⇒ X² +5X -1800 =0

We get a = 1, b = 5 and c = −1800

Applying quadratic formula 

⇒

⇒ 

⇒ 

⇒ x = 40, −45

Since speed of train cannot be in negative. Therefore, we discard x = −45

Therefore, speed of train = 40 km/hr

TIME/WORK

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9. Two water taps together can fill a tank in 9(3/8) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Solution: Let the time taken by the smaller pipe to fill the tank = x hr.

Time taken by the larger pipe = (x – 10) hr

Part of tank filled by smaller pipe in 1 hour = 1/x

Part of tank filled by larger pipe in 1 hour = 1/(x – 10)

As given, the tank can be filled in 9 3/8 = 75/8 hours by both the pipes together.

Therefore,

1/x + 1/x-10 = 8/75

(x-10+x)/x(x-10) = 8/75

⇒ 2x-10/x(x-10) = 8/75

⇒ 75(2x – 10) = 8x2 – 80x

⇒ 150x – 750 = 8x2 – 80x

⇒ 8x2 – 230x +750 = 0

⇒ 8x2 – 200x – 30x + 750 = 0

⇒ 8x(x – 25) -30(x – 25) = 0

⇒ (x – 25)(8x -30) = 0

⇒ x = 25, 30/8

Time taken by the smaller pipe cannot be 30/8 = 3.75 hours, as the time taken by the larger pipe will become negative, which is logically not possible.

Therefore, time taken individually by the smaller pipe and the larger pipe will be 25 and 25 – 10 =15 hours respectively.

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PERIMETER AND AREA

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11. Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.

Solutions: Let the sides of the two squares be x m and y m.

Therefore, their perimeter will be 4x and 4y respectively

And area of the squares will be x2 and y2 respectively.

Given,

4x – 4y = 24

x – y = 6

x = y + 6

Also, x2 + y2 = 468

⇒ (6 + y2) + y2 = 468

⇒ 36 + y2 + 12y + y2 = 468

⇒ 2y2 + 12y + 432 = 0

⇒ y2 + 6y – 216 = 0

⇒ y2 + 18y – 12y – 216 = 0

⇒ y(y +18) -12(y + 18) = 0

⇒ (y + 18)(y – 12) = 0

⇒ y = -18, 12

As we know, the side of a square cannot be negative.

Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m.

YOU ALL HAVE LEARNT ABOUT NATURE OF ROOTS IN YOUR PREVIOUS BLOGS!!

LET'S REVISE.

READ AND UNDERSTAND

Add caption

LET'S SOLVE Q1 (I.II) OF EX 4.4

MAKE A NOTE OF THIS

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

(i) 2x² – 3x + 5 = 0 

First.we will  find a =  2…..  ,b  = -3.....,   c =…..5.

then, Discriminant = b² - 4ac

      (-3)² - 4(2)(5)

      9 -40

     -31

since, D <0 span="">

so, roots are (real/equal/imaginary)

(ii)3x² – 4 √3 x + 4 = 0 

First.we will  find a =  3…..  ,b  = -  4 √3.....,   c =…..4.

then, Discriminant = b² - 4ac

      (-4 √3)² - 4(3)(4)

      48 -48

     0

since D=0 , roots are real and equal

so , roots are given by -b/2a  = -(-  4 √3)/ 2(3)  = 4 √3/6  

= 2√3/3
so roots are  2√3/3,2√3/3

   HOME WORK

EX 4.3 Q10 *refer to the video link given above

EX 4.4 Q1(III)

TAKE CARE AND KEEP SAFE!!