Students often refer to Kerala Syllabus 10th Standard Maths Textbook Solutions Chapter 9 Polynomials and Equations Questions and Answers Notes Pdf to clear their doubts.
SSLC Maths Chapter 9 Polynomials and Equations Questions and Answers
Polynomials and Equations Class 10 Questions and Answers Kerala State Syllabus
SCERT Class 10 Maths Chapter 9 Polynomials and Equations Solutions
Class 10 Maths Chapter 9 Kerala Syllabus – Multiplications
(Textbook Page No. 190)
Question 1.
Do the multiplications below in your head. Write the general principle from each set as an algebraic identity:
(i) 43 × 47
(ii) 63 × 67
(iii) 103 × 107
Answer:
(i) 43 × 47
= (40 + 3)(40 + 7)
= (40 × 40) + (40 × 7) + (3 × 40) + (3 × 7)
= 1600 + 40(7 + 3) + 21
= 1600 + 400 + 21
= 2021
(ii) 63 × 67
= (60 + 3)(60 + 7)
= (60 × 60) + (60 × 7) + (3 × 60) + (3 × 7)
= 3600 + 60 × (7 + 3) + 21
= 3600 + 600 + 21
= 4221
(iii) 103 × 107
= (100 + 3)(100 + 7)
= (100 × 100) + (100 × 7) + (3 × 100) + (3 × 7)
= 10000 + 100 × (7 + 3) + 21
= 10000 + 1000 + 21
= 11021
For any number x, (x + 3)(x + 7) = x2 + (3 + 7)x + 21 = x2 + 10x + 21
Question 2.
(i) 51 × 52
(ii) 81 × 82
(iii) 301 × 302
Answer:
(i) 51 × 52
= (50 + 1)(50 + 2)
= (50 × 50) + (50 × 2) + (1 × 50) + (1 × 2)
= 2500 + 50 × (1 + 2) + 2
= 2500 + 150 + 2
= 2652
(ii) 81 × 82
= (80 + 1)(80 + 2)
= (80 × 80) + (80 × 2) + (1 × 80) + (1 × 2)
= 6400 + 80 × (1 + 2) + 2
= 6400 + 240 + 2
= 6642
(iii) 301 × 302
= (300 + 1)(300 + 2)
= (300 × 300) + (300 × 2) + (1 × 300) + (1 × 2)
= 90000 + 300 × (1 + 2) + 2
= 90000 + 900 + 2
= 90902
For any number x, (x + 1)(x + 2) = x2 + (1 + 2)x + 2 = x2 + 3x + 2
SCERT Class 10 Maths Chapter 9 Solutions – Polynomial Multiplication
(Textbook Page No. 193)
Question 1.
Find the following products.
(i) (x + 2)(x + 5)
Answer:
(x + 2)(x + 5)
= x2 + (2 + 5) × x + 2 × 5
= x2 + 7x + 10
(ii) (x + 2)(x – 5)
Answer:
(x + 2)(x – 5)
= (x + 2)(x + (- 5))
= x2 + (2 + (-5)) × x + 2(-5)
= x2 – 3x – 10
(iii) (x – 2)(x + 5)
Answer:
(x – 2) (x + 5)
= (x + (-2))(x + 5)
= x2 + ((-2) + 5) × x + (-2) × 5
= x2 + 3x – 10
(iv) (x – 2)(x – 5)
Answer:
(x – 2)(x – 5)
= (x + (-2))(x + (-5))
= x2 + ((-2) + (-5)) × x + (-2)(-5)
= x2 – 7x + 10
Class 10 Maths Kerala Syllabus Chapter 9 Solutions – Polynomial Factors
(Textbook Page No. 195)
Question 1.
Write the second-degree polynomials below as the product of two first-degree polynomials:
(i) x2 + x – 6
Answer:
Take x2 + x – 6
(x + a)(x + b) = x2 + (a + b)x + ab = (x + a)(x + b)
x2 + x – 6 = x2 + (a + b)x + ab
⇒ a + b = 1, ab = -6
So, we have to find two numbers with a sum of 1 and a product of -6.
Thus, the numbers are -2 and 3.
Hence, x2 + x – 6 = (x + (-2)) (x + 3)
⇒ x2 + x – 6 = (x – 2)(x + 3)
(ii) x2 – x – 6
Answer:
Take x2 – x – 6
(x + a)(x + b) = x2 + (a + b)x + ab = (x + a)(x + b)
⇒ x2 – x – 6 = x2 + (a + b)x + ab
⇒ a + b = -1 , ab = -6
So, we have to find two numbers with a sum of -1 and a product of -6.
Thus, the numbers are 2 and -3.
Hence, x2 – x – 6 = (x + 2)(x + (-3))
⇒ x2 – x – 6 = (x + 2)(x – 3)
(iii) x2 + 7x + 6
Answer:
Take x2 + 7x + 6
(x + a)(x + b) = x2 + (a + b)x + ab = (x + a)(x + b)
⇒ x2 + 7x + 6 = x2 + (a + b)x + ab
⇒ a + b = 1, ab = 6
So, we have to find two numbers with a sum of 7 and a product of 6.
Thus, the numbers are 1 and 6.
Hence, x2 + 7x + 6 = (x + 1)(x + 6)
(iv) x2 – 7x + 6
Answer:
Take x2 – 7x + 6
(x + a) (x + b) = x2 + (a + b)x + ab = (x + a)(x + b)
⇒ x2 – 7x + 6 = x2 + (a + b)x + ab
⇒ a + b = -7, ab = 6
So, we have to find two numbers with a sum of -7 and a product of 6.
Thus, the numbers are -1 and -6.
Hence, x2 – 7x + 6 = (x + (-1))(x + (-6))
⇒ x2 – 7x + 6 = (x – 1)(x – 6)
In all these, we found the numbers in the first-degree polynomials simply by mental computation.
This may not always be possible.
In that type of problem, first find a-b using the identity (a-b)2 = (a + b)2 – 4ab, then by using a + b and a – b find the numbers a and b.
(Textbook Page No. 197)
Question 1.
Write the polynomials below as the product of two first-degree factors:
(i) x2 + 30x + 221
Answer:
Take x2 + 30x + 221 = (x + a)(x + b)
x2 + (a + b)x + ab = (x + a)(x + b)
⇒ x2 + 30x + 221 = x2 + (a + b)x + ab
⇒ a + b = 30, ab = 221
So, we have to find two numbers with a sum of 30 and a product of 221
(a – b)2 = (a + b)2 – 4ab
⇒ (a – b)2 = 302 – 4 × 221
⇒ (a – b)2 = 900 – 884 = 16
⇒ a – b = ±4
If we take a – b = 4, a + b = 30
⇒ a = \(\frac {1}{2}\)(30 + 4)
= \(\frac {1}{2}\) × 34
= 17
b = \(\frac {1}{2}\)(30 – 4)
= \(\frac {1}{2}\) × 26
= 13
If we take a – b = -4, a + b = 30
⇒ a = \(\frac {1}{2}\)(30 +(-4))
= \(\frac {1}{2}\) × 26
= 13
b = \(\frac {1}{2}\)(30 – (-4))
= \(\frac {1}{2}\) × 34
= 17
Using this, x2 + 30x + 221 = (x + 13)(x + 17)
(ii) x2 + 4x – 221
Answer:
Take x2 + 4x – 221 = (x + a)(x + b)
x2 + (a + b)x + ab = (x + a)(x + b)
⇒ x2 + 4x – 221 = x2 + (a + b)x + ab
⇒ a + b = 4, ab = -221
So, we have to find two numbers with a sum of 4 and a product of -221.
(a – b)2 = (a + b)2 – 4ab
⇒ (a – b)2 = 42 – 4 × (-221)
⇒ (a – b)2 = 16 + 884 = 900
⇒ a – b = ±30
If we take a – b = 30
a + b = 4, a – b = 30
⇒ a = \(\frac {1}{2}\)(4 + 30)
= \(\frac {1}{2}\) × 34 = 17
b = \(\frac {1}{2}\)(4 – 30)
= \(\frac {1}{2}\) × (-26)
= -13
If we take a – b = -30,
a + b = 4, a – b = -30
⇒ a = \(\frac {1}{2}\)(4 + (-30))
= \(\frac {1}{2}\) × (-26)
= -13
b = \(\frac {1}{2}\)(4 – (-30))
= \(\frac {1}{2}\) × 34
= 17
Using this, x2 + 4x – 221 = (x + (-13))(x + 17)
⇒ x2 + 4x – 221 = (x – 13)(x + 17)
(iii) x2 + x – 156
Answer:
Take x2 + x – 156 = (x + a)(x + b)
x2 + (a + b)x + ab = (x + a)(x + b)
⇒ x2 + x – 156 = x2 + (a + b)x + ab
⇒ a + b = 1, ab = -156
So, we have to find two numbers with a sum of 1 and a product of -156.
(a – b)2 = (a + b)2 – 4ab
⇒ (a – b)2 = 12 – 4 × (-156)
⇒ (a – b)2 = 1 + 624 = 625
⇒ a – b = ±25
If we take a – b = -25,
a + b = 1, a – b = -25
⇒ a = \(\frac {1}{2}\)(1 + 25)
= \(\frac {1}{2}\) × 26
= 13
b = \(\frac {1}{2}\)(1 – 25)
= \(\frac {1}{2}\) × (-24)
= -12
If we take a – b = -25
a + b = 1, a – b = -25
⇒ a = \(\frac {1}{2}\)(1 + (-25))
= \(\frac {1}{2}\) × (-24)
= -12
b = \(\frac {1}{2}\)(1 – (-25))
= \(\frac {1}{2}\) × 26
= 13
Using this, x2 + x – 156 = (x + 13) (x + (-12))
⇒ x2 + x – 156 = (x + 13) (x – 12)
SSLC Maths Chapter 9 Questions and Answers – Factors and Solutions
(Textbook Page No. 202)
Question 1.
One side of a rectangle is 2 metres longer than the other, and its area is 48 square metres. What are the lengths of its sides?
Answer:
If we denote the length of the shorter side as x metres, then the length of the longer side is x + 2 metres,
Area = x(x + 2) = 48
⇒ x2 + 2x = 48
⇒ x2 + 2x – 48 = 0
x2 + 2x – 48 = (x + a)(x + b) = x2 + (a + b)x + ab
⇒ a + b = 2, ab = -48
(a – b)2 = (a + b)2 – 4ab
⇒ (a – b)2 = 22 – 4 × (-48)
⇒ (a – b)2 = 4 + 192 = 196
⇒ a – b = ±14
Take a – b = 14,
a + b = 2, a – b = 14
⇒ a = \(\frac {1}{2}\)(2 + 14)
= \(\frac {1}{2}\) × 16
= 8
b = \(\frac {1}{2}\)(2 – 14)
= \(\frac {1}{2}\) × (-12)
= -6
Take a – b = -14,
a + b = 2, a – b = -14
⇒ a = \(\frac {1}{2}\)(2 + (-14))
= \(\frac {1}{2}\) × (-12)
= -6
b = \(\frac {1}{2}\)(2 – (-14))
= \(\frac {1}{2}\) × 16
= 8
Using this, x2 + 2x – 48 = (x + 8 )(x + (-6))
⇒ x2 + 2x – 48 = (x + 8)(x – 6)
x2 + 2x – 48 0
⇒ (x + 8) (x – 6) = 0
⇒ x = -8 or x = 6
x is the length of the side of the rectangle, so it cannot be a negative number.
So, the length of the shorter side of the rectangle is 6 metres.
The length of the longer side is 6 + 2 = 8 metres.
Question 2.
One side of a right triangle is one centimetre less than twice the length of the side perpendicular to it; the hypotenuse is one centimetre more than twice this side. What are the lengths of the sides?
Answer:
If we denote the length of one side of a right triangle as x metres,
Perpendicular side = 2x – 1
Hypotenuse = 2x + 1
Using Pythagoras’ theorem,
Hypotenuse2 = Base2 + Alttitude2
⇒ (2x + 1)2 = x2 + (2x – 1)2
⇒ 4x2 + 4x + 1 = x2 + 4x2 – 4x + 1
⇒ x2 – 8x = 0
⇒ x(x – 8) = 0
⇒ x = 0 or x = 8
x is the length of the side of the triangle, so it cannot be a negative number.
So, the length of the side = 8 metres.
Perpendicular side = 2 × 8 – 1 = 15 metres
Hypotenuse = 2 × 8 + 1 = 17 metres
Question 3.
The sum of the consecutive natural numbers from 1 up to which number is 300?
Answer:
Let the sum of the consecutive natural numbers from 1 up to n be 300.
The sum of the consecutive natural numbers from 1 up to n = \(\frac{n(n+1)}{2}\)
⇒ \(\frac{n(n+1)}{2}\) = 300
⇒ \(\frac{n^2+n}{2}\) = 300
⇒ n2 + n = 600
⇒ n2 + n – 600 = 0
n2 + n – 600 = (n + a)(n + b) = n2 + (a + b)n + ab
⇒ a + b = 1, ab = -600
(a – b)2 = (a + b)2 – 4ab
⇒ (a – b)2 = 12 – 4 × (-600)
⇒ (a – b)2 = 1 + 2400 = 2401
⇒ a – b = ±49
Take a – b = 49
a + b = 1, a – b = 49
⇒ a = \(\frac {1}{2}\)(1 – 49)
= \(\frac {1}{2}\) × 50
= 25
b = \(\frac {1}{2}\)(1 – 49)
= \(\frac {1}{2}\) × (-48)
= -24
Take a – b = -49
a + b = 1, a – b = -49
⇒ a = \(\frac {1}{2}\)(1 – 49)
= \(\frac {1}{2}\) × (-48)
= -24
b = \(\frac {1}{2}\)(1 – (-49))
= \(\frac {1}{2}\) × (1 + 49)
= 25
Using this,
n2 + n – 600 = (n + 25)(n – 24)
n2 + n – 600 = 0
⇒ (n + 25)(n – 24) = 0
⇒ n = -25 or n = 24
n is the count of numbers, so it cannot be negative.
So, n = 24
The sum of the consecutive natural numbers from 1 up to 24 is 300.
Question 4.
At what points does the graph of the polynomial x2 – 2x – 1 cross the x-axis?
Answer:
If the graph of the polynomial x2 – 2x – 1 cross the x-axis, x2 – 2x – 1 = 0
x2 – 2x – 1 = (x + a)(x + b) = x2 + (a + b)x + ab
⇒ a + b = -2, ab = -1
(a – b)2 = (a + b)2 – 4ab
⇒ (a – b)2 = (-2)2 – 4 × (-1)
⇒ (a – b)2 = 4 + 4 = 8
⇒ a – b = ±√8
⇒ a – b = ±2√2
Take a – b = 2√2
a + b = -2, a – b = 2√2
⇒ a = \(\frac {1}{2}\)(-2 + 2√2)
= \(\frac {1}{2}\) × 2(-1 + √2)
= √2 – 1
b = \(\frac {1}{2}\)(-2 – 2√2)
= \(\frac {1}{2}\) × 2(-1 – √2)
= -1 – √2
Using this, x2 – 2x – 1 = (x + √2 – 1)(x + (-1 – √2))
x2 – 2x – 1 = 0
⇒ (x + √2 – 1)(x + (-1 – √2)) = 0
⇒ x = 1 – √2 or x = 1 + √2
The points at which the graph of the polynomial x2 – 2x – 1 crosses the x-axis are (1 – √2, 0), (1 + √2, 0)
Kerala Syllabus Class 10 Maths Chapter 1 Solutions – General Solution
(Textbook Page No. 206)
Question 1.
Find the points at which the graphs of the polynomials given below cross the x-axis:
(i) 2x2 – 7x – 1
(ii) 2x2 + 7x – 1
(iii) 9x2 + 12x + 4
Answer:
(i) 2x2 – 7x – 1 = 0
a = 2, b = -7, c = -1
So,
Point at which the graph crosses the x-axis: (-3, 0)
Question 2.
The perimeter of a rectangle is 42 metres, and its diagonal is 15 metres. What are the lengths of its sides?
Answer:
If we denote the length of the rectangle as x metres,
Perimeter = 42 metres
⇒ 2 (length + breadth) = 42
⇒ length + breadth = 21
⇒ x + breadth = 21
⇒ breadth = 21 – x
Hypotenuse = 15 metres
⇒ x2 + (21 – x)2 = 152
⇒ x2 + 212 – 42x + x2 = 152
⇒ 2x2 – 42x + 441 = 225
⇒ 2x2 – 42x + 216 = 0
a = 2, b = -42, c = 216
So, x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
If we take the length of the rectangle as 12 metres, then the breadth = 21 – 12 = 9 metres.
If we take the length of the rectangle as 9 metres, then the breadth = 21 – 9 = 12 metres.
Question 3.
From a rectangular sheet of paper, two squares are cut off as shown below:
The area of the remaining part is 27 square centimetres. What is the length of the shorter side of the rectangle?
Answer:
Let the side of the cut-off squares be x, then the length of the remaining rectangle is 15 – 2x
Breadth = x
Area of the remaining part = 27 square metres
⇒ x(15 – 2x) = 27
⇒ 15x – 2x2 = 27
⇒ 2x2 – 15x + 27 = 0
a = 2, b = -15, c = 27
⇒ x = \(\frac{15+3}{4}=\frac{18}{4}=\frac{9}{2}\) = 4.5 or x = \(\frac{15-3}{4}\) = 3
The length of the other side of the rectangle = 4.5 metres or 3 metres.
Question 4.
How many terms of the arithmetic sequence, 1, 5, 9,… starting from the first, are to be added to get 91?
Answer:
Let the first n terms of an arithmetic sequence be added to get the sum 91.
Then nth term = dn + (f – d)
= 4n + (1 – 4)
= 4n – 3
Sum of first n terms = \(\frac {n}{2}\)(4n – 3 + 1)
= \(\frac {n}{2}\)(4n – 2)
= \(\frac {n}{2}\) × 2(2n – 1)
= n(2n – 1)
= 2n2 – n
So, 2n2 – n = 91
⇒ 2n2 – n – 91 = 0
a = 2, b = -1, c = -91
Since n is the number of terms, it must be a positive integer.
∴ n = 7
∴ The sum of the first 7 terms of the arithmetic sequence is 91.
Question 5.
A rectangle is to be made by bending a 28-centimeter-long rod.
(i) Can a rectangle of diagonal 8 centimetres be made?
(ii) How about a rectangle of diagonal 10 centimetres?
(iii) And a rectangle of diagonal 14 centimetres?
Calculate the lengths of the sides of those rectangles that can be made as above.
Answer:
(i) Perimeter of the rectangle = 28 centimetres
Let the length of the rectangle be x.
⇒ 2(length + breadth) = 28
⇒ length + breadth = 14
⇒ x + breadth = 14
⇒ breadth = 14 – x
If the length of the diagonal is 8 centimeters,
x2 + (14 – x)2 = 82
⇒ x2 + 142 – 28x + x2 = 82
⇒ 2x2 – 28x + 196 = 64
⇒ 2x2 – 28x + 132 = 0
a = 2, b = -28, c = 132
Hence,
∴ A rectangle with a perimeter of 28 cm and a diagonal of 8 cm cannot be made.
(i) If the length of the diagonal is 10 centimeters,
x2 + (14 – x)2 = 102
⇒ x2 + 142 – 28x + x2 = 102
⇒ 2x2 – 28x + 196 = 100
⇒ 2x2 – 28x + 96 = 0
a = 2, b = -28, c = 96
Hence,
If the length of the rectangle is 8 cm, then the breadth = 14 – 8 = 6 cm
If the length of the rectangle is 6 cm, then the breadth = 14 – 6 = 8 cm
A rectangle diagonal of 10 cm can be made.
(iii) If the length of the diagonal is 14 centimeters,
x2 + (14 – x)2 = 142
⇒ x2 + 142 – 28x + x2 = 142
⇒ 2x2 – 28x + 196 = 196
⇒ 2x2 – 28x = 0
⇒ 2x(x – 14) = 0
⇒ x = 0 or x – 14 = 0
⇒ x = 0 or x = 14
x is the length of a side of the rectangle, so x ≠ 0
Hence, the length of the rectangle = 14 cm
Breadth = 14 – 14 = 0 cm
A rectangle diagonal of 14 cm cannot be made.
Polynomials and Equations Class 10 Notes Pdf
Class 10 Maths Chapter 9 Polynomials and Equations Notes Kerala Syllabus
Introduction
A polynomial is a mathematical expression consisting of indeterminates (also called variables) and coefficients, that involves only the operations of addition, subtraction, multiplication, and exponentiation to nonnegative integer powers, and has a finite number of terms.
Example: 2x2 – 3x + 1
→ Each part of a polynomial is called a term.
For example, 2x2 is a term.
→ Variables in polynomials are symbols, often letters, that represent unknown or changing values within a mathematical expression.
→ In the polynomial 2x2 – 3x + 1, x is a variable.
→ Addition, subtraction, multiplication, division, etc, are mathematical operations.
→ An equation with power two is known as a second-degree equation (or Quadratic equation).
→ General form of a second-degree equation is ax² + bx + c = 0
→ The solutions of the equation ax2 + bx + c = 0 are given by x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
→ To multiply a sum by a sum, each number in one sum is to be multiplied by each number in the other sum, and the products added.
For any four numbers x, y, u, v.
(x + y)(u + v) = xu + xv + yu + yv
→ The product of the first degree polynomials x + a and x + b is the second degree polynomial x2 + (a + b)x + ab.
For any numbers x, a, b,
(x + a)(x + b) = x2 + (a + b)x + ab
(x + a)(x – b) = x2 + (a – b)x – ab
(x – a)(x – b) = x2 – (a + b)x + ab
→ We can write a second-degree polynomial as the product of two first-degree polynomials.
Multiplications
To multiply a sum by a sum, each number in one sum is to be multiplied by each number in the other sum, and the products added.
If we wrote this as an algebraic equation
For any four numbers x, y, u, v
(x + y)(u + v) = xu + xv + yu + yv
Polynomial Multiplication
For any numbers x, a, b,
(x + a)(x + b) = x2 + (a + b)x + ab
In terms of polynomials,
The product of the first degree polynomials x + a and x + b is the second degree polynomial x2 + (a + b)x + ab
(x – a)(x – b) = x2 – (a + b)x + ab
Polynomial Factors
We can write a second-degree polynomial as the product of two first-degree polynomials.
For example, let us look at how we can write x2 + 5x + 6 as the product of two first-degree polynomials:
Take x2 + 5x + 6 = (x + a)(x + b)
x2 + (a + b)x + ab = (x + a)(x + b)
⇒ x2 + 5x + 6 = x2 + (a + b)x + ab
⇒ a + b = 5, ab = 6
So, we have to find two numbers with a sum of 5 and a product of 6.
Thus, the numbers are 2 and 3.
Hence, x2 + 5x + 6 = (x + 2)(x + 3)
Factors and Solutions
One side of a rectangle is 3 metres longer than the other side, and its area is 270 square metres. What are the lengths of the sides?
Answer:
If we denote the length of the shorter side as x metres, then the length of the longer side is x + 3 metres.
Area = x(x + 3) = 270 square metres.
The problem can be stated using algebra as
To get x2 + 3x = 270
What number should we take as x?
Factorize x2 + 3x – 270
x2 + 3x – 270 = (x + a) (x + b) = x2 + (a + b)x + ab
⇒ a + b = 3, ab = -270
Using identity (a – b)2 = (a + b)2 – 4ab, find a – b.
⇒ (a – b)2 = 32 – 4 × (-270)
⇒ (a – b)2 = 9 + 1080 = 1089
To find a – b, calculate the square root of 1089.
1089 = 32 × 112 = (3 × 11)2 = 332
⇒ a – b = ±33
Take a – b = 33,
a + b = 3, a – b = 33
⇒ a = \(\frac {1}{2}\) × (3 + 33)
= \(\frac {1}{2}\) × 36
= 18
b = \(\frac {1}{2}\) × (3-33)
= \(\frac {1}{2}\) × (-30)
= -15
Using this, x2 + 3x – 270 = (x + 18)(x + (-15))
⇒ x2 + 3x – 270 = (x + 18)(x – 15)
x2 + 3x – 270 = 0
⇒ To get (x + 18)(x – 15) = 0
What number should we take as x?
⇒ x = -18 or x = 15
x is the length of the side of the rectangle, so it cannot be a negative number.
So, the length of the shorter side of the rectangle is 15 metres.
The length of the longer side is 15 + 3 = 18 metres.
One added to a number gives the square of the number. What is the number?
Answer:
Let the number be x,
x + 1 = x2
⇒ To get x2 – x – 1 = 0
What number should we take as x?
Factorize x2 – x – 1
x2 – x – 1 = (x + a)(x + b) = x2 + (a + b)x + ab
a + b = -1, ab = -1
(a – b)2 = (a + b)2 – 4ab
⇒ (a – b)2 = (-1)2 – 4 × (-1)
⇒ (a – b)2 = 1 + 4 = 5
⇒ a – b = ±√5
Take a – b = √5,
a + b = -1, a – b = √5
1 added to either of \(-\frac{\sqrt{5}-1}{2}, \frac{1+\sqrt{5}}{2}\) gives its square.
General Solution
Any second-degree polynomial can be written in the form ax² + bx + c.
So any second-degree equation can be written in the form ax² + bx + c = 0
⇒ x2 + \(\frac {b}{a}\)x + \(\frac {c}{a}\) = 0
⇒ x2 + \(\frac {b}{a}\)x + \(\frac {c}{a}\) = (x + p)(x + q) = x2 + (p + q)x + pq
⇒ p + q = \(\frac {b}{a}\) and pq = \(\frac {c}{a}\)
By using the identity (p-q)2 = (p + q)2 – 4pq, find p-q. Using p-q and p + q, find p, q.
So, we can write the solutions of the equation ax² + bx + c = 0
x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)