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
9x2 + 4 = 0
⟹(3x)2 - (2i)2 = 0 [∵ i2 = – 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)2 - 4i2 = 0
⟹ (2x – 3 + 4i) (2x – 3 - 4i) = 0
⟹ (2x – 3 + 4i) = 0 or (2x – 3 – 4i) = 0
Quadratic Equations – Exercise 14.1 – Q.5
x2 + x + 1 = 0
Now, completing the squares, we get
Quadratic Equations – Exercise 14.1 – Q.6
4x2 + 1 = 0
⟹(2x)2 - i2 = 0 [ ∵ i2 = – 1]
⟹(2x + i)(2x – i) = 0
⟹ either 2x + i = 0 or 2x – i = 0
Quadratic Equations – Exercise 14.1 – Q.7
x2 - 4x + 7 = 0
We will apply discriminant rule,
Where D = b2 - 4ac =(-4)2 - 4.1.7 = – 12
from (A)
.png)
Quadratic Equations – Exercise 14.1 – Q.8
x2 + 2x + 2 = 0
We will apply discriminant rule,
Where D = b2 - 4ac
= 22 - 4.1.2
= 4 – 8
= – 4
Quadratic Equations – Exercise 14.1 – Q.9
5x2 - 6x + 2 = 0
We will apply discriminant rule,
Where D = b2 - 4ac
=(-b)2 - 4.5.2
= 36 – 40
= – 4
from (A)
.png)
Quadratic Equations – Exercise 14.1 – Q.10
21x2+ 9x + 1 = 0
Comparing the given equation with the general form
ax2 + bx + c = 0, we get a = 21, b = 9, c = 1
Substituting a and b in,
Quadratic Equations – Exercise 14.1 – Q.11
x2 - x + 1 = 0
We will apply discriminant rule,
Where D = b2 - 4ac
=(-1)2 - 4.1.1
=1 – 4
= – 3
from (A)
.png)
Quadratic Equations – Exercise 14.1 – Q.12
x2 + x + 1 = 0
We will apply discriminant rule,
Where D = b2 - 4ac
=12 - 4.1.1
=1 – 4
= – 3
from (A)
.png)
Quadratic Equations – Exercise 14.1 – Q.13
17x2 - 8x + 1 = 0
We will apply discriminant rule,
Where D = b2 - 4ac
=(-8)2 - 4.17.1
= 64 – 68
= – 4
from (A)
.png)
Quadratic Equations – Exercise 14.1 – Q.14
27x2 - 10x + 1 = 0
We will apply discriminant rule,
Where D = b2 - 4ac
= (-10)2- 4.27.1
=100 – 108
= – 8
from (A)
.png)
View courses by askIITians
Design classes One-on-One in your own way with Top IITians/Medical Professionals
Click Here Know More
Complete Self Study Package designed by Industry Leading Experts
Click Here Know More
Live 1-1 coding classes to unleash the Creator in your Child
Click Here Know More
a Complete All-in-One Study package Fully Loaded inside a Tablet!
Click Here Know MoreAsk a Doubt
Get your questions answered by the expert for free