In earlier classes, we have studied quadratic equations with real coefficients and real roots only. In this chapter, we shall study quadratic equations with real coefficients and complex roots. We shall also discuss quadratic equations with complex coefficients and their solutions in the complex number system. For a better understanding of the concepts, students can solve the exercise wise problems using the solutions, which are developed by our expert faculty team at BYJU’S. Students aspiring to secure high marks in their examination are advised to practice the solutions on a regular basis. RD Sharma Class 11 Maths Solutions pdf can be downloaded from the links provided below.
Chapter 14 – Quadratic Equations contains two exercises and the RD Sharma Solutions present in this page provide solutions to the questions present in each exercise. Now, let us have a look at the concepts discussed in this chapter.
- Some useful definitions and results.
- Quadratic equations with real coefficients.
- Quadratic equations with complex coefficients.
Download the pdf of RD Sharma Solutions for Class 11 Maths Chapter 14 – Quadratic Equations
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EXERCISE 14.1 PAGE NO: 14.5
Solve the following quadratic equations by factorization method only:
1. x2 + 1 = 0
Solution:
Given: x2Â + 1 = 0
We know, i2 = –1 ⇒ 1 = –i2
By substituting 1 = –i2 in the above equation, we get
x2 – i2 = 0
[By using the formula, a2 – b2 = (a + b) (a – b)](x + i) (x – i) = 0
x + i = 0 or x – i = 0
x = –i or x = i
∴ The roots of the given equation are i, -i
2. 9x2Â + 4 = 0
Solution:
Given: 9x2Â + 4 = 0
9x2 + 4 × 1 = 0
We know, i2 = –1 ⇒ 1 = –i2
By substituting 1 = –i2 in the above equation, we get
So,
9x2 + 4(–i2) = 0
9x2 – 4i2 = 0
(3x)2 – (2i)2 = 0
[By using the formula, a2 – b2 = (a + b) (a – b)](3x + 2i) (3x – 2i) = 0
3x + 2i = 0 or 3x – 2i = 0
3x = –2i or 3x = 2i
x = -2i/3 or x = 2i/3
∴ The roots of the given equation are 2i/3, -2i/3
3. x2Â + 2x + 5 = 0
Solution:
Given: x2Â + 2x + 5 = 0
x2Â + 2x + 1 + 4 = 0
x2Â + 2(x) (1) + 12Â + 4 = 0
(x + 1)2 + 4 = 0 [since, (a + b)2 = a2 + 2ab + b2]
(x + 1)2 + 4 × 1 = 0
We know, i2 = –1 ⇒ 1 = –i2
By substituting 1 = –i2 in the above equation, we get
(x + 1)2 + 4(–i2) = 0
(x + 1)2 – 4i2 = 0
(x + 1)2 – (2i)2 = 0
[By using the formula, a2 – b2 = (a + b) (a – b)](x + 1 + 2i)(x + 1 – 2i) = 0
x + 1 + 2i = 0 or x + 1 – 2i = 0
x = –1 – 2i or x = –1 + 2i
∴ The roots of the given equation are -1+2i, -1-2i
4. 4x2 – 12x + 25 = 0
Solution:
Given: 4x2 – 12x + 25 = 0
4x2 – 12x + 9 + 16 = 0
(2x)2 – 2(2x)(3) + 32 + 16 = 0
(2x – 3)2 + 16 = 0 [Since, (a + b)2 = a2 + 2ab + b2]
(2x – 3)2 + 16 × 1 = 0
We know, i2 = –1 ⇒ 1 = –i2
By substituting 1 = –i2 in the above equation, we get
(2x – 3)2 + 16(–i2) = 0
(2x – 3)2 – 16i2 = 0
(2x – 3)2 – (4i)2 = 0
[By using the formula, a2 – b2 = (a + b) (a – b)](2x – 3 + 4i) (2x – 3 – 4i) = 0
2x – 3 + 4i = 0 or 2x – 3 – 4i = 0
2x = 3 – 4i or 2x = 3 + 4i
x = 3/2 – 2i or x = 3/2 + 2i
∴ The roots of the given equation are 3/2 + 2i, 3/2 – 2i
5. x2Â + x + 1 = 0
Solution:
Given: x2Â + x + 1 = 0
x2 + x + ¼ + ¾ = 0
x2 + 2 (x) (1/2) + (1/2)2 + ¾ = 0
(x + 1/2)2 + ¾ = 0 [Since, (a + b)2 = a2 + 2ab + b2]
(x + 1/2)2 + ¾ × 1 = 0
We know, i2 = –1 ⇒ 1 = –i2
By substituting 1 = –i2 in the above equation, we get
(x + ½)2 + ¾ (-1)2 = 0
(x + ½)2 + ¾ i2 = 0
(x + ½)2 – (√3i/2)2 = 0
[By using the formula, a2 – b2 = (a + b) (a – b)](x + ½ + √3i/2) (x + ½ – √3i/2) = 0
(x + ½ + √3i/2) = 0 or (x + ½ – √3i/2) = 0
x = -1/2 – √3i/2 or x = -1/2 + √3i/2
∴ The roots of the given equation are -1/2 + √3i/2, -1/2 – √3i/2
6. 4x2Â + 1 = 0
Solution:
Given: 4x2Â + 1 = 0
We know, i2 = –1 ⇒ 1 = –i2
By substituting 1 = –i2 in the above equation, we get
4x2 – i2 = 0
(2x)2 – i2 = 0
[By using the formula, a2 – b2 = (a + b) (a – b)](2x + i) (2x – i) = 0
2x + i = 0 or 2x – i = 0
2x = –i or 2x = i
x = -i/2 or x = i/2
∴ The roots of the given equation are i/2, -i/2
7. x2 – 4x + 7 = 0
Solution:
Given: x2 – 4x + 7 = 0
x2 – 4x + 4 + 3 = 0
x2 – 2(x) (2) + 22 + 3 = 0
(x – 2)2 + 3 = 0 [Since, (a – b)2 = a2 – 2ab + b2]
(x – 2)2 + 3 × 1 = 0
We know, i2 = –1 ⇒ 1 = –i2
By substituting 1 = –i2 in the above equation, we get
(x – 2)2 + 3(–i2) = 0
(x – 2)2 – 3i2 = 0
(x – 2)2 – (√3i)2 = 0
[By using the formula, a2 – b2 = (a + b) (a – b)](x – 2 + √3i) (x – 2 – √3i) = 0
(x – 2 + √3i) = 0 or (x – 2 – √3i) = 0
x = 2 – √3i or x = 2 + √3i
x = 2 ± √3i
∴ The roots of the given equation are 2 ± √3i
8. x2Â + 2x + 2 = 0
Solution:
Given: x2Â + 2x + 2 = 0
x2Â + 2x + 1 + 1 = 0
x2Â + 2(x)(1) + 12Â + 1 = 0
(x + 1)2 + 1 = 0 [∵ (a + b)2 = a2 + 2ab + b2]
We know, i2 = –1 ⇒ 1 = –i2
By substituting 1 = –i2 in the above equation, we get
(x + 1)2 + (–i2) = 0
(x + 1)2 – i2 = 0
(x + 1)2 – (i)2 = 0
[By using the formula, a2 – b2 = (a + b) (a – b)](x + 1 + i) (x + 1 – i) = 0
x + 1 + i = 0 or x + 1 – i = 0
x = –1 – i or x = –1 + i
x = -1 ± i
∴ The roots of the given equation are -1 ± i
9. 5x2 – 6x + 2 = 0
Solution:
Given: 5x2 – 6x + 2 = 0
We shall apply discriminant rule,
Where, x = (-b ±√(b2 – 4ac))/2a
Here, a = 5, b = -6, c = 2
So,
x = (-(-6) ±√(-62 – 4 (5)(2)))/ 2(5)
= (6 ± √(36-40))/10
= (6 ± √(-4))/10
= (6 ± √4(-1))/10
We have i2 = –1
By substituting –1 = i2 in the above equation, we get
x = (6 ± √4i2)/10
= (6 ± 2i)/10
= 2(3±i)/10
= (3±i)/5
x = 3/5 ± i/5
∴ The roots of the given equation are 3/5 ± i/5
10. 21x2Â + 9x + 1 = 0
Solution:
Given: 21x2Â + 9x + 1 = 0
We shall apply discriminant rule,
Where, x = (-b ±√(b2 – 4ac))/2a
Here, a = 21, b = 9, c = 1
So,
x = (-9 ±√(92 – 4 (21)(1)))/ 2(21)
= (-9 ± √(81-84))/42
= (-9 ± √(-3))/42
= (-9 ± √3(-1))/42
We have i2 = –1
By substituting –1 = i2 in the above equation, we get
x = (-9 ± √3i2)/42
= (-9 ± √(√3i)2/42
= (-9 ± √3i)/42
= -9/42 ± √3i/42
= -3/14 ± √3i/42
∴ The roots of the given equation are -3/14 ± √3i/42
11. x2 – x + 1 = 0
Solution:
Given: x2 – x + 1 = 0
x2 – x + ¼ + ¾ = 0
x2 – 2 (x) (1/2) + (1/2)2 + ¾ = 0
(x – 1/2)2 + ¾ = 0 [Since, (a + b)2 = a2 + 2ab + b2]
(x – 1/2)2 + ¾ × 1 = 0
We know, i2 = –1 ⇒ 1 = –i2
By substituting 1 = –i2 in the above equation, we get
(x – ½)2 + ¾ (-1)2 = 0
(x – ½)2 + ¾ (-i)2 = 0
(x – ½)2 – (√3i/2)2 = 0
[By using the formula, a2 – b2 = (a + b) (a – b)](x – ½ + √3i/2) (x – ½ – √3i/2) = 0
(x – ½ + √3i/2) = 0 or (x – ½ – √3i/2) = 0
x = 1/2 – √3i/2 or x = 1/2 + √3i/2
∴ The roots of the given equation are 1/2 + √3i/2, 1/2 – √3i/2
12. x2Â + x + 1 = 0
Solution:
Given: x2Â + x + 1 = 0
x2 + x + ¼ + ¾ = 0
x2 + 2 (x) (1/2) + (1/2)2 + ¾ = 0
(x + 1/2)2 + ¾ = 0 [Since, (a + b)2 = a2 + 2ab + b2]
(x + 1/2)2 + ¾ × 1 = 0
We know, i2 = –1 ⇒ 1 = –i2
By substituting 1 = –i2 in the above equation, we get
(x + ½)2 + ¾ (-1)2 = 0
(x + ½)2 + ¾ i2 = 0
(x + ½)2 – (√3i/2)2 = 0
[By using the formula, a2 – b2 = (a + b) (a – b)](x + ½ + √3i/2) (x + ½ – √3i/2) = 0
(x + ½ + √3i/2) = 0 or (x + ½ – √3i/2) = 0
x = -1/2 – √3i/2 or x = -1/2 + √3i/2
∴ The roots of the given equation are -1/2 + √3i/2, -1/2 – √3i/2
13. 17x2 – 8x + 1 = 0
Solution:
Given: 17x2 – 8x + 1 = 0
We shall apply discriminant rule,
Where, x = (-b ±√(b2 – 4ac))/2a
Here, a = 17, b = -8, c = 1
So,
x = (-(-8) ±√(-82 – 4 (17)(1)))/ 2(17)
= (8 ± √(64-68))/34
= (8 ± √(-4))/34
= (8 ± √4(-1))/34
We have i2 = –1
By substituting –1 = i2 in the above equation, we get
x = (8 ± √(2i)2)/34
= (8 ± 2i)/34
= 2(4±i)/34
= (4±i)/17
x = 4/17 ± i/17
∴ The roots of the given equation are 4/17 ± i/17
EXERCISE 14.2 PAGE NO: 14.13
1. Solving the following quadratic equations by factorization method:
(i) x2 + 10ix – 21 = 0
(ii) x2 + (1 – 2i)x – 2i = 0
(iii) x2 – (2√3 + 3i) x + 6√3i = 0
(iv) 6x2 – 17ix – 12 = 0
Solution:
(i) x2 + 10ix – 21 = 0
Given: x2 + 10ix – 21 = 0
x2 + 10ix – 21 × 1 = 0
We know, i2 = –1 ⇒ 1 = –i2
By substituting 1 = –i2 in the above equation, we get
x2 + 10ix – 21(–i2) = 0
x2Â + 10ix + 21i2Â = 0
x2Â + 3ix + 7ix + 21i2Â = 0
x(x + 3i) + 7i(x + 3i) = 0
(x + 3i) (x + 7i) = 0
x + 3i = 0 or x + 7i = 0
x = –3i or –7i
∴ The roots of the given equation are –3i, –7i
(ii) x2 + (1 – 2i)x – 2i = 0
Given: x2 + (1 – 2i)x – 2i = 0
x2 + x – 2ix – 2i = 0
x(x + 1) – 2i(x + 1) = 0
(x + 1) (x – 2i) = 0
x + 1 = 0 or x – 2i = 0
x = –1 or 2i
∴ The roots of the given equation are –1, 2i
(iii) x2 – (2√3 + 3i) x + 6√3i = 0
Given: x2 – (2√3 + 3i) x + 6√3i = 0
x2 – (2√3x + 3ix) + 6√3i = 0
x2 – 2√3x – 3ix + 6√3i = 0
x(x – 2√3) – 3i(x – 2√3) = 0
(x – 2√3) (x – 3i) = 0
(x – 2√3) = 0 or (x – 3i) = 0
x = 2√3 or x = 3i
∴ The roots of the given equation are 2√3, 3i
(iv) 6x2 – 17ix – 12 = 0
Given: 6x2 – 17ix – 12 = 0
6x2 – 17ix – 12 × 1 = 0
We know, i2 = –1 ⇒ 1 = –i2
By substituting 1 = –i2 in the above equation, we get
6x2 – 17ix – 12(–i2) = 0
6x2 – 17ix + 12i2 = 0
6x2 – 9ix – 8ix + 12i2 = 0
3x(2x – 3i) – 4i(2x – 3i) = 0
(2x – 3i) (3x – 4i) = 0
2x – 3i = 0 or 3x – 4i = 0
2x = 3i or 3x = 4i
x = 3i/2 or x = 4i/3
∴ The roots of the given equation are 3i/2, 4i/3
2. Solve the following quadratic equations:
(i) x2 – (3√2 + 2i) x + 6√2i = 0
(ii) x2 – (5 – i) x + (18 + i) = 0
(iii) (2 + i)x2 – (5- i)x + 2 (1 – i) = 0
(iv) x2 – (2 + i)x – (1 – 7i) = 0
(v) ix2 – 4x – 4i = 0
(vi) x2 + 4ix – 4 = 0
(vii) 2x2 + √15ix – i = 0Â
(viii) x2 – x + (1 + i) = 0
(ix) ix2 – x + 12i = 0
(x) x2 – (3√2 – 2i)x – √2i = 0
(xi) x2 – (√2 + i)x + √2i = 0
(xii) 2x2 – (3 + 7i)x + (9i – 3) = 0
Solution:
(i) x2 – (3√2 + 2i) x + 6√2i = 0
Given: x2 – (3√2 + 2i) x + 6√2i = 0
x2 – (3√2x + 2ix) + 6√2i = 0
x2 – 3√2x – 2ix + 6√2i = 0
x(x – 3√2) – 2i(x – 3√2) = 0
(x – 3√2) (x – 2i) = 0
(x – 3√2) = 0 or (x – 2i) = 0
x = 3√2 or x = 2i
∴ The roots of the given equation are 3√2, 2i
(ii) x2 – (5 – i) x + (18 + i) = 0
Given: x2 – (5 – i) x + (18 + i) = 0
We shall apply discriminant rule,
Where, x = (-b ±√(b2 – 4ac))/2a
Here, a = 1, b = -(5-i), c = (18+i)
So,
We can write 48 + 14i = 49 – 1 + 14i
So,
48 + 14i = 49 + i2 + 14i [∵ i2 = –1]
= 72Â + i2Â + 2(7)(i)
= (7 + i)2 [Since, (a + b)2 = a2 + b2 + 2ab]
By using the result 48 + 14i = (7 + i) 2, we get
x = 2 + 3i or 3 – 4i
∴ The roots of the given equation are 3 – 4i, 2 + 3i
(iii) (2 + i)x2 – (5- i)x + 2 (1 – i) = 0
Given: (2 + i)x2 – (5- i)x + 2 (1 – i) = 0
We shall apply discriminant rule,
Where, x = (-b ±√(b2 – 4ac))/2a
Here, a = (2+i), b = -(5-i), c = 2(1-i)
So,
We have i2 = –1
By substituting –1 = i2 in the above equation, we get
x = (1 – i) or 4/5 – 2i/5
∴ The roots of the given equation are (1 – i), 4/5 – 2i/5
(iv) x2 – (2 + i)x – (1 – 7i) = 0
Given: x2 – (2 + i)x – (1 – 7i) = 0
We shall apply discriminant rule,
Where, x = (-b ±√(b2 – 4ac))/2a
Here, a = 1, b = -(2+i), c = -(1-7i)
So,
We can write 7 – 24i = 16 – 9 – 24i
7 – 24i = 16 + 9(–1) – 24i
= 16 + 9i2 – 24i [∵ i2 = –1]
= 42 + (3i)2 – 2(4) (3i)
= (4 – 3i)2 [∵ (a – b)2 = a2 – b2 + 2ab]
By using the result 7 – 24i = (4 – 3i)2, we get
x = 3 – i or -1 + 2i
∴ The roots of the given equation are (-1 + 2i), (3 – i)
(v) ix2 – 4x – 4i = 0
Given: ix2 – 4x – 4i = 0
ix2 + 4x(–1) – 4i = 0 [We know, i2 = –1]
So by substituting –1 = i2 in the above equation, we get
ix2 + 4xi2 – 4i = 0
i(x2 + 4ix – 4) = 0
x2 + 4ix – 4 = 0
x2 + 4ix + 4(–1) = 0
x2 + 4ix + 4i2 = 0 [Since, i2 = –1]
x2Â + 2ix + 2ix + 4i2Â = 0
x(x + 2i) + 2i(x + 2i) = 0
(x + 2i) (x + 2i) = 0
(x + 2i)2Â = 0
x + 2i = 0
x = –2i, -2i
∴ The roots of the given equation are –2i, –2i
(vi) x2 + 4ix – 4 = 0
Given: x2 + 4ix – 4 = 0
x2 + 4ix + 4(–1) = 0 [We know, i2 = –1]
So by substituting –1 = i2 in the above equation, we get
x2Â + 4ix + 4i2Â = 0
x2Â + 2ix + 2ix + 4i2Â = 0
x(x + 2i) + 2i(x + 2i) = 0
(x + 2i) (x + 2i) = 0
(x + 2i)2Â = 0
x + 2i = 0
x = –2i, -2i
∴ The roots of the given equation are –2i, –2i
(vii) 2x2 + √15ix – i = 0Â
Given: 2x2 + √15ix – i = 0Â
We shall apply discriminant rule,
Where, x = (-b ±√(b2 – 4ac))/2a
Here, a = 2, b = √15i, c = -i
So,
We can write 15 – 8i = 16 – 1 – 8i
15 – 8i = 16 + (–1) – 8i
= 16 + i2 – 8i [∵ i2 = –1]
= 42 + (i)2 – 2(4)(i)
= (4 – i)2 [Since, (a – b)2 = a2 – b2 + 2ab]
By using the result 15 – 8i = (4 – i)2, we get
∴ The roots of the given equation are [1+ (4 – √15)i/4] , [-1 -(4 + √15)i/4]
(viii) x2 – x + (1 + i) = 0
Given: x2 – x + (1 + i) = 0
We shall apply discriminant rule,
Where, x = (-b ±√(b2 – 4ac))/2a
Here, a = 1, b = -1, c = (1+i)
So,
We can write 3 + 4i = 4 – 1 + 4i
3 + 4i = 4 + i2 + 4i [∵ i2 = –1]
= 22Â + i2Â + 2(2) (i)
= (2 + i)2 [Since, (a + b)2 = a2 + b2 + 2ab]
By using the result 3 + 4i = (2 + i)2, we get
x = 2i/2 or (2 – 2i)/2
x = i or 2(1-i)/2
x = i or (1 – i)
∴ The roots of the given equation are (1-i), i
(ix) ix2 – x + 12i = 0
Given: ix2 – x + 12i = 0
ix2 + x(–1) + 12i = 0 [We know, i2 = –1]
so by substituting –1 = i2 in the above equation, we get
ix2Â + xi2Â + 12i = 0
i(x2Â + ix + 12) = 0
x2Â + ix + 12 = 0
x2 + ix – 12(–1) = 0
x2 + ix – 12i2 = 0 [Since, i2 = –1]
x2 – 3ix + 4ix – 12i2 = 0
x(x – 3i) + 4i(x – 3i) = 0
(x – 3i) (x + 4i) = 0
x – 3i = 0 or x + 4i = 0
x = 3i or –4i
∴ The roots of the given equation are -4i, 3i
(x) x2 – (3√2 – 2i)x – √2i = 0
Given: x2 – (3√2 – 2i)x – √2i = 0
We shall apply discriminant rule,
Where, x = (-b ±√(b2 – 4ac))/2a
Here, a = 1, b = -(3√2 – 2i), c = –√2i
So,
(xi) x2 – (√2 + i)x + √2i = 0
Given: x2 – (√2 + i)x + √2i = 0
x2 – (√2x + ix) + √2i = 0
x2 – √2x – ix + √2i = 0
x(x – √2) – i(x – √2) = 0
(x – √2) (x – i) = 0
(x – √2) = 0 or (x – i) = 0
x = √2 or x = i
∴ The roots of the given equation are i, √2
(xii) 2x2 – (3 + 7i)x + (9i – 3) = 0
Given: 2x2 – (3 + 7i)x + (9i – 3) = 0
We shall apply discriminant rule,
Where, x = (-b ±√(b2 – 4ac))/2a
Here, a = 2, b = -(3 + 7i), c = (9i – 3)
So,
We can write 16 + 30i = 25 – 9 + 30i
16 + 30i = 25 + 9(–1) + 30i
= 25 + 9i2 + 30i [∵ i2 = –1]
= 52Â + (3i)2Â + 2(5)(3i)
= (5 + 3i)2 [∵ (a + b)2 = a2 + b2 + 2ab]
By using the result 16 + 30i = (5 + 3i)2, we get
