RD Sharma Solutions Commerce Maths Class 11 Class 11 Solutions Chapter - Quadratic

 Page 1


14. Quadratic Equations
Exercise 14.1
1. Question
Solve the following quadratic equations by factorization method
x
2
 + 1 = 0
Answer
Given x
2
 + 1 = 0
We have i
2
 = –1 ? 1 = –i
2
By substituting 1 = –i
2
 in the above equation, we get
x
2
 – i
2
 = 0
? (x + i)(x – i) = 0 [? a
2
 – b
2
 = (a + b)(a – b)]
? x + i = 0 or x – i = 0
? x = –i or x = i
? x = ±i
Thus, the roots of the given equation are ±i.
2. Question
Solve the following quadratic equations by factorization method
9x
2
 + 4 = 0
Answer
Given 9x
2
 + 4 = 0
? 9x
2
 + 4 × 1 = 0
We have i
2
 = –1 ? 1 = –i
2
By substituting 1 = –i
2
 in the above equation, we get
9x
2
 + 4(–i
2
) = 0
? 9x
2
 – 4i
2
 = 0
? (3x)
2
 – (2i)
2
 = 0
? (3x + 2i)(3x – 2i) = 0 [? a
2
 – b
2
 = (a + b)(a – b)]
? 3x + 2i = 0 or 3x – 2i = 0
? 3x = –2i or 3x = 2i
Thus, the roots of the given equation are .
3. Question
Solve the following quadratic equations by factorization method
Page 2


14. Quadratic Equations
Exercise 14.1
1. Question
Solve the following quadratic equations by factorization method
x
2
 + 1 = 0
Answer
Given x
2
 + 1 = 0
We have i
2
 = –1 ? 1 = –i
2
By substituting 1 = –i
2
 in the above equation, we get
x
2
 – i
2
 = 0
? (x + i)(x – i) = 0 [? a
2
 – b
2
 = (a + b)(a – b)]
? x + i = 0 or x – i = 0
? x = –i or x = i
? x = ±i
Thus, the roots of the given equation are ±i.
2. Question
Solve the following quadratic equations by factorization method
9x
2
 + 4 = 0
Answer
Given 9x
2
 + 4 = 0
? 9x
2
 + 4 × 1 = 0
We have i
2
 = –1 ? 1 = –i
2
By substituting 1 = –i
2
 in the above equation, we get
9x
2
 + 4(–i
2
) = 0
? 9x
2
 – 4i
2
 = 0
? (3x)
2
 – (2i)
2
 = 0
? (3x + 2i)(3x – 2i) = 0 [? a
2
 – b
2
 = (a + b)(a – b)]
? 3x + 2i = 0 or 3x – 2i = 0
? 3x = –2i or 3x = 2i
Thus, the roots of the given equation are .
3. Question
Solve the following quadratic equations by factorization method
x
2
 + 2x + 5 = 0
Answer
Given x
2
 + 2x + 5 = 0
? x
2
 + 2x + 1 + 4 = 0
? x
2
 + 2(x)(1) + 1
2
 + 4 = 0
? (x + 1)
2
 + 4 = 0 [? (a + b)
2
 = a
2
 + 2ab + b
2
]
? (x + 1)
2
 + 4 × 1 = 0
We have i
2
 = –1 ? 1 = –i
2
By substituting 1 = –i
2
 in the above equation, we get
(x + 1)
2
 + 4(–i
2
) = 0
? (x + 1)
2
 – 4i
2
 = 0
? (x + 1)
2
 – (2i)
2
 = 0
? (x + 1 + 2i)(x + 1 – 2i) = 0 [? a
2
 – b
2
 = (a + b)(a – b)]
? x + 1 + 2i = 0 or x + 1 – 2i = 0
? x = –1 – 2i or x = –1 + 2i
? x = –1 ± 2i
Thus, the roots of the given equation are –1 ± 2i.
4. Question
Solve the following quadratic equations by factorization method
4x
2
 – 12x + 25 = 0
Answer
Given 4x
2
 – 12x + 25 = 0
? 4x
2
 – 12x + 9 + 16 = 0
? (2x)
2
 – 2(2x)(3) + 3
2
 + 16 = 0
? (2x – 3)
2
 + 16 = 0 [? (a + b)
2
 = a
2
 + 2ab + b
2
]
? (2x – 3)
2
 + 16 × 1 = 0
We have i
2
 = –1 ? 1 = –i
2
By substituting 1 = –i
2
 in the above equation, we get
(2x – 3)
2
 + 16(–i
2
) = 0
? (2x – 3)
2
 – 16i
2
 = 0
? (2x – 3)
2
 – (4i)
2
 = 0
Since a
2
 – b
2
 = (a + b)(a – b), we get
(2x – 3 + 4i)(2x – 3 – 4i) = 0
? 2x – 3 + 4i = 0 or 2x – 3 – 4i = 0
? 2x = 3 – 4i or 2x = 3 + 4i
Page 3


14. Quadratic Equations
Exercise 14.1
1. Question
Solve the following quadratic equations by factorization method
x
2
 + 1 = 0
Answer
Given x
2
 + 1 = 0
We have i
2
 = –1 ? 1 = –i
2
By substituting 1 = –i
2
 in the above equation, we get
x
2
 – i
2
 = 0
? (x + i)(x – i) = 0 [? a
2
 – b
2
 = (a + b)(a – b)]
? x + i = 0 or x – i = 0
? x = –i or x = i
? x = ±i
Thus, the roots of the given equation are ±i.
2. Question
Solve the following quadratic equations by factorization method
9x
2
 + 4 = 0
Answer
Given 9x
2
 + 4 = 0
? 9x
2
 + 4 × 1 = 0
We have i
2
 = –1 ? 1 = –i
2
By substituting 1 = –i
2
 in the above equation, we get
9x
2
 + 4(–i
2
) = 0
? 9x
2
 – 4i
2
 = 0
? (3x)
2
 – (2i)
2
 = 0
? (3x + 2i)(3x – 2i) = 0 [? a
2
 – b
2
 = (a + b)(a – b)]
? 3x + 2i = 0 or 3x – 2i = 0
? 3x = –2i or 3x = 2i
Thus, the roots of the given equation are .
3. Question
Solve the following quadratic equations by factorization method
x
2
 + 2x + 5 = 0
Answer
Given x
2
 + 2x + 5 = 0
? x
2
 + 2x + 1 + 4 = 0
? x
2
 + 2(x)(1) + 1
2
 + 4 = 0
? (x + 1)
2
 + 4 = 0 [? (a + b)
2
 = a
2
 + 2ab + b
2
]
? (x + 1)
2
 + 4 × 1 = 0
We have i
2
 = –1 ? 1 = –i
2
By substituting 1 = –i
2
 in the above equation, we get
(x + 1)
2
 + 4(–i
2
) = 0
? (x + 1)
2
 – 4i
2
 = 0
? (x + 1)
2
 – (2i)
2
 = 0
? (x + 1 + 2i)(x + 1 – 2i) = 0 [? a
2
 – b
2
 = (a + b)(a – b)]
? x + 1 + 2i = 0 or x + 1 – 2i = 0
? x = –1 – 2i or x = –1 + 2i
? x = –1 ± 2i
Thus, the roots of the given equation are –1 ± 2i.
4. Question
Solve the following quadratic equations by factorization method
4x
2
 – 12x + 25 = 0
Answer
Given 4x
2
 – 12x + 25 = 0
? 4x
2
 – 12x + 9 + 16 = 0
? (2x)
2
 – 2(2x)(3) + 3
2
 + 16 = 0
? (2x – 3)
2
 + 16 = 0 [? (a + b)
2
 = a
2
 + 2ab + b
2
]
? (2x – 3)
2
 + 16 × 1 = 0
We have i
2
 = –1 ? 1 = –i
2
By substituting 1 = –i
2
 in the above equation, we get
(2x – 3)
2
 + 16(–i
2
) = 0
? (2x – 3)
2
 – 16i
2
 = 0
? (2x – 3)
2
 – (4i)
2
 = 0
Since a
2
 – b
2
 = (a + b)(a – b), we get
(2x – 3 + 4i)(2x – 3 – 4i) = 0
? 2x – 3 + 4i = 0 or 2x – 3 – 4i = 0
? 2x = 3 – 4i or 2x = 3 + 4i
Thus, the roots of the given equation are .
5. Question
Solve the following quadratic equations by factorization method
x
2
 + x + 1 = 0
Answer
Given x
2
 + x + 1 = 0
 [? (a + b)
2
 = a
2
 + 2ab + b
2
]
We have i
2
 = –1 ? 1 = –i
2
By substituting 1 = –i
2
 in the above equation, we get
 [? a
2
 – b
2
 = (a + b)(a – b)]
Thus, the roots of the given equation are .
6. Question
Solve the following quadratics
4x
2
 + 1 = 0
Answer
Page 4


14. Quadratic Equations
Exercise 14.1
1. Question
Solve the following quadratic equations by factorization method
x
2
 + 1 = 0
Answer
Given x
2
 + 1 = 0
We have i
2
 = –1 ? 1 = –i
2
By substituting 1 = –i
2
 in the above equation, we get
x
2
 – i
2
 = 0
? (x + i)(x – i) = 0 [? a
2
 – b
2
 = (a + b)(a – b)]
? x + i = 0 or x – i = 0
? x = –i or x = i
? x = ±i
Thus, the roots of the given equation are ±i.
2. Question
Solve the following quadratic equations by factorization method
9x
2
 + 4 = 0
Answer
Given 9x
2
 + 4 = 0
? 9x
2
 + 4 × 1 = 0
We have i
2
 = –1 ? 1 = –i
2
By substituting 1 = –i
2
 in the above equation, we get
9x
2
 + 4(–i
2
) = 0
? 9x
2
 – 4i
2
 = 0
? (3x)
2
 – (2i)
2
 = 0
? (3x + 2i)(3x – 2i) = 0 [? a
2
 – b
2
 = (a + b)(a – b)]
? 3x + 2i = 0 or 3x – 2i = 0
? 3x = –2i or 3x = 2i
Thus, the roots of the given equation are .
3. Question
Solve the following quadratic equations by factorization method
x
2
 + 2x + 5 = 0
Answer
Given x
2
 + 2x + 5 = 0
? x
2
 + 2x + 1 + 4 = 0
? x
2
 + 2(x)(1) + 1
2
 + 4 = 0
? (x + 1)
2
 + 4 = 0 [? (a + b)
2
 = a
2
 + 2ab + b
2
]
? (x + 1)
2
 + 4 × 1 = 0
We have i
2
 = –1 ? 1 = –i
2
By substituting 1 = –i
2
 in the above equation, we get
(x + 1)
2
 + 4(–i
2
) = 0
? (x + 1)
2
 – 4i
2
 = 0
? (x + 1)
2
 – (2i)
2
 = 0
? (x + 1 + 2i)(x + 1 – 2i) = 0 [? a
2
 – b
2
 = (a + b)(a – b)]
? x + 1 + 2i = 0 or x + 1 – 2i = 0
? x = –1 – 2i or x = –1 + 2i
? x = –1 ± 2i
Thus, the roots of the given equation are –1 ± 2i.
4. Question
Solve the following quadratic equations by factorization method
4x
2
 – 12x + 25 = 0
Answer
Given 4x
2
 – 12x + 25 = 0
? 4x
2
 – 12x + 9 + 16 = 0
? (2x)
2
 – 2(2x)(3) + 3
2
 + 16 = 0
? (2x – 3)
2
 + 16 = 0 [? (a + b)
2
 = a
2
 + 2ab + b
2
]
? (2x – 3)
2
 + 16 × 1 = 0
We have i
2
 = –1 ? 1 = –i
2
By substituting 1 = –i
2
 in the above equation, we get
(2x – 3)
2
 + 16(–i
2
) = 0
? (2x – 3)
2
 – 16i
2
 = 0
? (2x – 3)
2
 – (4i)
2
 = 0
Since a
2
 – b
2
 = (a + b)(a – b), we get
(2x – 3 + 4i)(2x – 3 – 4i) = 0
? 2x – 3 + 4i = 0 or 2x – 3 – 4i = 0
? 2x = 3 – 4i or 2x = 3 + 4i
Thus, the roots of the given equation are .
5. Question
Solve the following quadratic equations by factorization method
x
2
 + x + 1 = 0
Answer
Given x
2
 + x + 1 = 0
 [? (a + b)
2
 = a
2
 + 2ab + b
2
]
We have i
2
 = –1 ? 1 = –i
2
By substituting 1 = –i
2
 in the above equation, we get
 [? a
2
 – b
2
 = (a + b)(a – b)]
Thus, the roots of the given equation are .
6. Question
Solve the following quadratics
4x
2
 + 1 = 0
Answer
Given 4x
2
 + 1 = 0
We have i
2
 = –1 ? 1 = –i
2
By substituting 1 = –i
2
 in the above equation, we get
4x
2
 – i
2
 = 0
? (2x)
2
 – i
2
 = 0
? (2x + i)(2x – i) = 0 [? a
2
 – b
2
 = (a + b)(a – b)]
? 2x + i = 0 or 2x – i = 0
? 2x = –i or 2x = i
Thus, the roots of the given equation are .
7. Question
Solve the following quadratics
x
2
 – 4x + 7 = 0
Answer
Given x
2
 – 4x + 7 = 0
? x
2
 – 4x + 4 + 3 = 0
? x
2
 – 2(x)(2) + 2
2
 + 3 = 0
? (x – 2)
2
 + 3 = 0 [? (a – b)
2
 = a
2
 – 2ab + b
2
]
? (x – 2)
2
 + 3 × 1 = 0
We have i
2
 = –1 ? 1 = –i
2
By substituting 1 = –i
2
 in the above equation, we get
(x – 2)
2
 + 3(–i
2
) = 0
? (x – 2)
2
 – 3i
2
 = 0
Since a
2
 – b
2
 = (a + b)(a – b), we get
Thus, the roots of the given equation are .
8. Question
Solve the following quadratics
Page 5


14. Quadratic Equations
Exercise 14.1
1. Question
Solve the following quadratic equations by factorization method
x
2
 + 1 = 0
Answer
Given x
2
 + 1 = 0
We have i
2
 = –1 ? 1 = –i
2
By substituting 1 = –i
2
 in the above equation, we get
x
2
 – i
2
 = 0
? (x + i)(x – i) = 0 [? a
2
 – b
2
 = (a + b)(a – b)]
? x + i = 0 or x – i = 0
? x = –i or x = i
? x = ±i
Thus, the roots of the given equation are ±i.
2. Question
Solve the following quadratic equations by factorization method
9x
2
 + 4 = 0
Answer
Given 9x
2
 + 4 = 0
? 9x
2
 + 4 × 1 = 0
We have i
2
 = –1 ? 1 = –i
2
By substituting 1 = –i
2
 in the above equation, we get
9x
2
 + 4(–i
2
) = 0
? 9x
2
 – 4i
2
 = 0
? (3x)
2
 – (2i)
2
 = 0
? (3x + 2i)(3x – 2i) = 0 [? a
2
 – b
2
 = (a + b)(a – b)]
? 3x + 2i = 0 or 3x – 2i = 0
? 3x = –2i or 3x = 2i
Thus, the roots of the given equation are .
3. Question
Solve the following quadratic equations by factorization method
x
2
 + 2x + 5 = 0
Answer
Given x
2
 + 2x + 5 = 0
? x
2
 + 2x + 1 + 4 = 0
? x
2
 + 2(x)(1) + 1
2
 + 4 = 0
? (x + 1)
2
 + 4 = 0 [? (a + b)
2
 = a
2
 + 2ab + b
2
]
? (x + 1)
2
 + 4 × 1 = 0
We have i
2
 = –1 ? 1 = –i
2
By substituting 1 = –i
2
 in the above equation, we get
(x + 1)
2
 + 4(–i
2
) = 0
? (x + 1)
2
 – 4i
2
 = 0
? (x + 1)
2
 – (2i)
2
 = 0
? (x + 1 + 2i)(x + 1 – 2i) = 0 [? a
2
 – b
2
 = (a + b)(a – b)]
? x + 1 + 2i = 0 or x + 1 – 2i = 0
? x = –1 – 2i or x = –1 + 2i
? x = –1 ± 2i
Thus, the roots of the given equation are –1 ± 2i.
4. Question
Solve the following quadratic equations by factorization method
4x
2
 – 12x + 25 = 0
Answer
Given 4x
2
 – 12x + 25 = 0
? 4x
2
 – 12x + 9 + 16 = 0
? (2x)
2
 – 2(2x)(3) + 3
2
 + 16 = 0
? (2x – 3)
2
 + 16 = 0 [? (a + b)
2
 = a
2
 + 2ab + b
2
]
? (2x – 3)
2
 + 16 × 1 = 0
We have i
2
 = –1 ? 1 = –i
2
By substituting 1 = –i
2
 in the above equation, we get
(2x – 3)
2
 + 16(–i
2
) = 0
? (2x – 3)
2
 – 16i
2
 = 0
? (2x – 3)
2
 – (4i)
2
 = 0
Since a
2
 – b
2
 = (a + b)(a – b), we get
(2x – 3 + 4i)(2x – 3 – 4i) = 0
? 2x – 3 + 4i = 0 or 2x – 3 – 4i = 0
? 2x = 3 – 4i or 2x = 3 + 4i
Thus, the roots of the given equation are .
5. Question
Solve the following quadratic equations by factorization method
x
2
 + x + 1 = 0
Answer
Given x
2
 + x + 1 = 0
 [? (a + b)
2
 = a
2
 + 2ab + b
2
]
We have i
2
 = –1 ? 1 = –i
2
By substituting 1 = –i
2
 in the above equation, we get
 [? a
2
 – b
2
 = (a + b)(a – b)]
Thus, the roots of the given equation are .
6. Question
Solve the following quadratics
4x
2
 + 1 = 0
Answer
Given 4x
2
 + 1 = 0
We have i
2
 = –1 ? 1 = –i
2
By substituting 1 = –i
2
 in the above equation, we get
4x
2
 – i
2
 = 0
? (2x)
2
 – i
2
 = 0
? (2x + i)(2x – i) = 0 [? a
2
 – b
2
 = (a + b)(a – b)]
? 2x + i = 0 or 2x – i = 0
? 2x = –i or 2x = i
Thus, the roots of the given equation are .
7. Question
Solve the following quadratics
x
2
 – 4x + 7 = 0
Answer
Given x
2
 – 4x + 7 = 0
? x
2
 – 4x + 4 + 3 = 0
? x
2
 – 2(x)(2) + 2
2
 + 3 = 0
? (x – 2)
2
 + 3 = 0 [? (a – b)
2
 = a
2
 – 2ab + b
2
]
? (x – 2)
2
 + 3 × 1 = 0
We have i
2
 = –1 ? 1 = –i
2
By substituting 1 = –i
2
 in the above equation, we get
(x – 2)
2
 + 3(–i
2
) = 0
? (x – 2)
2
 – 3i
2
 = 0
Since a
2
 – b
2
 = (a + b)(a – b), we get
Thus, the roots of the given equation are .
8. Question
Solve the following quadratics
x
2
 + 2x + 2 = 0
Answer
Given x
2
 + 2x + 2 = 0
? x
2
 + 2x + 1 + 1 = 0
? x
2
 + 2(x)(1) + 1
2
 + 1 = 0
? (x + 1)
2
 + 1 = 0 [? (a + b)
2
 = a
2
 + 2ab + b
2
]
We have i
2
 = –1 ? 1 = –i
2
By substituting 1 = –i
2
 in the above equation, we get
(x + 1)
2
 + (–i
2
) = 0
? (x + 1)
2
 – i
2
 = 0
? (x + 1)
2
 – (i)
2
 = 0
? (x + 1 + i)(x + 1 – i) = 0 [? a
2
 – b
2
 = (a + b)(a – b)]
? x + 1 + i = 0 or x + 1 – i = 0
? x = –1 – i or x = –1 + i
? x = –1 ± i
Thus, the roots of the given equation are –1 ± i.
9. Question
Solve the following quadratics
5x
2
 – 6x + 2 = 0
Answer
Given 5x
2
 – 6x + 2 = 0
Recall that the roots of quadratic equation ax
2
 + bx + c = 0, where a ? 0, are given by
Here, a = 5, b = –6 and c = 2
We have i
2
 = –1
By substituting –1 = i
2
 in the above equation, we get