Quadratic Equations | Complete NCERT WITH BACK EXERCISE in 1 Video | Class 10th Board

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00:00 - Introduction
02:00 - Exercise - 4.1
45:43 - Exercise - 4.2
01:46:03 - Exercise - 4.3
02:14:41 - Thank You !

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1:31:16-
Let the first integer be x.

The next consecutive positive integer will be x + 1.

According to the given question, the sum of squares of x and x + 1 is 365. i.e.,

x2 + ( x + 1)2 = 365

x2 +( x + 1)2 = 365

x2 + (x2 + 2x + 1) = 365 [ ∵ (a + b)2 = a2 + 2ab + b2]

2x2 + 2x + 1 = 365

2x2 + 2x + 1- 365 = 0

2x2 + 2x - 364 = 0

2(x2 + x - 182) = 0

x2 + x - 182 = 0

x2 + 14x - 13x - 182 = 0

x (x + 14) - 13 (x + 14) = 0

(x - 13) (x + 14) = 0

x - 13 = 0 and x + 14 = 0

x = 13 and x = - 14

1:35:22 -
In a right triangle, altitude is one of the sides.

Let the base be x cm.

The altitude will be (x - 7) cm.

We can now apply the Pythagoras theorem to the given right triangle.

Pythagoras theorem: Hypotenuse2 = (side 1)2 + (side 2)2

132 = x2 + (x - 7)2

132 = x2 + (x - 7)2

169 = x2 + x2 - 14x + 49

169 = 2x2 - 14x + 49

2x2 - 14x + 49 -169 = 0

2x2 - 14x - 120 = 0

(2x2 - 14x - 120) / 2 = 0

x2 - 7x - 60 = 0

x2 - 12x + 5x - 60 = 0

x(x - 12) + 5 (x - 12) = 0

(x + 5) (x - 12) = 0

x - 12 = 0 and x + 5 = 0

x = 12 and x = - 5

We know that the value of the base cannot be negative.

Therefore, Base = 12 cm, Altitude = 12 - 7 = 5 cm

2:00:49-
(ii) kx (x - 2) + 6 = 0

kx2 - 2kx + 6 = 0

a = k, b = - 2k, c = 6

b2 - 4ac = 0

(-2k)2 - 4(k)(6) = 0

4k2 - 24k = 0

4k (k - 6) = 0

k = 6 and k = 0

If we consider the value of k as 0, then the equation will no longer be quadratic.
or
Because A Quadratic equation has ax² + bx + c = 0 Conditions Where a ≠ 0 in this equation Kx² + 2Kx + 6 = 0 Where the values of K are K=0 and K=6 but According to the rule of Quadratic Equation a≠0 but here K=a so the value of k is 6 k=6 k≠0

Therefore, k = 6

2:13:56 -
Consider a rectangular park with length as 'l' and breadth as 'b' respectively.

Perimeter of a rectangle = 2(l + b) = 80 ....(1)

Area of a rectangle = l × b = 400 ....(2)

2(l + b) = 80

(l + b) = 40

l = 40 - b

Substituting the value of l = 40 - b in equation (2)

(40 - b)(b) = 400

40b - b2 = 400

40b - b2 - 400 = 0

b2 - 40b + 400 = 0

Let’s find the discriminant: b2 - 4ac

a = 1, b = - 40, c = 400

b2 - 4ac = (- 40)2 - 4(1)(400)

= 1600 - 1600

= 0

Since, the value of the discriminant is 0, thus we can have two equal and real roots.

Therefore, it is possible to design a rectangular park with the given condition.

x = [- b ± √(b2 - 4ac)] / 2a

= (- b ± 0) / 2a

= -(- 40) / 2(1)

= 40 / 2

= 20

So, breadth of the rectangle is b = 20 m and its length is l = 40 - b = 20 m

Note that the park will be square in shape with side length 20 m.

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Why k cannot be zero? 2:01:28
Because the general form of quadratic equation is ax²+bx+c=0 and where 'a' is not equal to zero and in that question kx(x-2) +6=0 "k" Means "a" Means a value is k and "a" is not equal to zero so that's why...!!!

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1:42:10 pr aapne galat kiya h 90 ko x^2 ke coddecient k sath multiply krege na ki x ke to iska shi hoga -180 naki -270...

[MehakSakoliya]

Standard form of quadratic equation is ax²+bx+c=0, where a is not equal to zero..

[mananchoure]

2:01:33 Because A Quadratic equation has ax² + bx + c = 0 Conditions Where a ≠ 0 in this equation Kx² + 2Kx + 6 = 0 Where the values of K are K=0 and K=6 but According to the rule of Quadratic Equation a≠0 but here K=a so the value of k is 6 k=6 k≠0

[unknown-id_]

2:01:33 k≠0 because, k = a which is coefficient of x² and a≠0(1:46:30) ∴ k≠0

[shankar_verma]

1:35:34 altitude= 5cm and base = 12cm.