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Chapter 14: Quadratic Equations – Exercise 14.1

Quadratic Equations – Exercise 14.1 – Q.1

x2 + 1 = 0

⟹ x2 + i2 = 0                [∵ i2 = – 1]

⟹ (x + i)(x - i) = 0       [a2 - b2 = (a + b)(a - b)]

⟹ x = i, -i

 

Quadratic Equations – Exercise 14.1 – Q.2

9x+ 4 = 0

⟹(3x)- (2i)= 0              [∵ i= – 1]

⟹ (3x + 2i)(3x – 2i) = 0

⟹ 3x + 2i = 0 or 3x = 2i = 0

 

Quadratic Equations – Exercise 14.1 – Q.3

x2 + 2x + 5 = 0

Now, completing the squares, we get

(x + 1)2 + 4 = 0

⟹ (x + 1)2 - 2i2 = 0

⟹ (x + 1 + 2i)(x + 1 - 2i) = 0

⟹ (x + 1 + 2i) = 0 or (x + 1 - 2i) = 0

∴ x = – 1 – 2i, – 1 + 2i

 

Quadratic Equations – Exercise 14.1 – Q.4

4x2 - 12x + 25 = 0

⟹ (2x – 3)- 4i= 0

 ⟹ (2x – 3 + 4i) (2x – 3 - 4i) = 0

⟹ (2x – 3 + 4i) = 0 or (2x – 3 – 4i) = 0

Quadratic Equations – Exercise 14.1 – Q.4
 

Quadratic Equations – Exercise 14.1 – Q.5

x+ x + 1 = 0

Now, completing the squares, we get

Quadratic Equations – Exercise 14.1 – Q.5

 

Quadratic Equations – Exercise 14.1 – Q.6

4x+ 1 = 0

⟹(2x)- i= 0 [ ∵ i= – 1]

⟹(2x + i)(2x – i) = 0

⟹ either 2x + i = 0 or 2x – i = 0

Quadratic Equations – Exercise 14.1 – Q.6

 

Quadratic Equations – Exercise 14.1 – Q.7

x- 4x + 7 = 0

We will apply discriminant rule,

Quadratic Equations – Exercise 14.1 – Q.7

Where D = b- 4ac =(-4)- 4.1.7 = – 12

from (A)

 

Quadratic Equations – Exercise 14.1 – Q.8

x+ 2x + 2 = 0

We will apply discriminant rule,

Quadratic Equations – Exercise 14.1 – Q.8

Where D = b- 4ac

= 2- 4.1.2

= 4 – 8

= – 4

 

Quadratic Equations – Exercise 14.1 – Q.9

5x- 6x + 2 = 0

We will apply discriminant rule,

Quadratic Equations – Exercise 14.1 – Q.9

Where D = b- 4ac

=(-b)- 4.5.2

= 36 – 40

= – 4

from (A)

 

Quadratic Equations – Exercise 14.1 – Q.10

21x2+ 9x + 1 = 0

Comparing the given equation with the general form

ax+ bx + c = 0, we get a = 21, b = 9, c = 1

Substituting a and b in,

Quadratic Equations – Exercise 14.1 – Q.10

 

Quadratic Equations – Exercise 14.1 – Q.11

x- x + 1 = 0

We will apply discriminant rule,

Quadratic Equations – Exercise 14.1 – Q.11

Where D = b- 4ac

=(-1)- 4.1.1

=1 – 4

= – 3

from (A)

 

Quadratic Equations – Exercise 14.1 – Q.12

x+ x + 1 = 0

We will apply discriminant rule,

Quadratic Equations – Exercise 14.1 – Q.12

Where D = b- 4ac

=1- 4.1.1

=1 – 4

= – 3

from (A)

 

Quadratic Equations – Exercise 14.1 – Q.13

17x- 8x + 1 = 0

We will apply discriminant rule,

Quadratic Equations – Exercise 14.1 – Q.13

Where D = b- 4ac

=(-8)- 4.17.1

= 64 – 68

= – 4

from (A)

 

Quadratic Equations – Exercise 14.1 – Q.14

27x- 10x + 1 = 0

We will apply discriminant rule,

Quadratic Equations – Exercise 14.1 – Q.14

Where D = b- 4ac

= (-10)2- 4.27.1

=100 – 108

= – 8

from (A)


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