Polynomials and Equations Class 10 Extra Questions Kerala State Syllabus Maths Chapter 9

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SSLC Maths Chapter 9 Polynomials and Equations Important Questions and Answers

Polynomials and Equations Class 10 Extra Questions Kerala Syllabus

Polynomials and Equations Class 10 Kerala Syllabus Extra Questions

Question 1.
At what points does the graph of the polynomial 4x² – 8x + 3 cross the x-axis?
(a) \(\left(\frac{3}{4}, 0\right),\left(\frac{1}{2}, 0\right)\)
(b) \(\left(\frac{3}{2}, 0\right),\left(\frac{1}{2}, 0\right)\)
(c) \(\left(\frac{3}{4}, 0\right),\left(\frac{1}{6}, 0\right)\)
(d) \(\left(\frac{7}{2}, 0\right),\left(\frac{3}{2}, 0\right)\)
Answer:
(b) \(\left(\frac{3}{2}, 0\right),\left(\frac{1}{2}, 0\right)\)
4x² – 8x + 3 = 0
a = 4, b = -8, c = 3

So, the points at which the graph of the polynomial 4x² – 8x + 3 crosses the x-axis are \(\left(\frac{3}{2}, 0\right),\left(\frac{1}{2}, 0\right)\)

Question 2.
The sum and product of the solutions of a second-degree polynomial are -10 and \(\frac {5}{2}\) respectively. Then the polynomial is:
(a) 2x² – 20x + 10
(b) 2x² – x + 5
(c) 2x² – 20x + 5
(d) x² – 20x + 5
Answer:
(c) 2x² – 20x + 5
Sum of the solutions is -10, and its product is \(\frac {5}{2}\)
a + b = -10, ab = \(\frac {5}{2}\)
Polynomial = x² + (a + b)x + ab
= x² – 10x + \(\frac {5}{2}\)
= 2x² – 20x + 5

Question 3.
Consider the following statements.
Statement (A): x² + 20x + 96 is the polynomial with the sum of solutions 8 and their product 12.
Statement (B): If a, b are the solutions of a polynomial, then the polynomial is x² + (a + b)x + ab.
(a) Statement A is true, Statement B is false.
(b) Statement B is true, Statement A is false.
(c) Both statements are true. Statement B is the reason for Statement A.
(d) Both statements are true. Statement B is not the reason for Statement A.
Answer:
(b) Statement B is true, Statement A is false.
Statement B is true.
The sum of solutions 8 and their product 12.
⇒ a + b = 8, ab = 12
Polynomial = x² + (a + b)x + ab = x² + 8x + 12
∴ Statement A is false.

Question 4.
Consider the following statements.
Statement (A): The area of a rectangle with sides (x + 5) and (x – 3) is a second-degree polynomial.
Statement (B): A second-degree polynomial can be written as the product of two first-degree polynomials.
(a) Statement A is true, Statement B is false.
(b) Statement B is true, Statement A is false.
(c) Both statements are true. Statement B is the reason for Statement A.
(d) Both statements are true. Statement B is not the reason for Statement A.
Answer:
(d) Both statements are true. Statement B is not the reason for Statement A.
Statement B is true.
The area of a rectangle with sides (x + 5) and (x – 3)
(x + 5)(x – 3) = x² + 5x – 3x – 15 = x² + 2x – 15
A second-degree polynomial Statement A is true.

Question 5.
Write the polynomial x2 + 5x – 84 as the product of two first-degree polynomials.
Answer:
x² + 5x – 84 = (x + a)(x + b) = x² + (a + b)x + ab
⇒ a + b = 5, ab = -84
⇒ (a – b)² = (a + b)² – 4ab
⇒ (a – b)² = 5² – 4 × (-84)
⇒ (a – b)² = 25 + 336
⇒ (a – b)² = 361
⇒ a – b = ±19
Take a – b = 19
a + b = 5, a – b = 19
⇒ a = \(\frac {1}{2}\)(5 + 19) = \(\frac {1}{2}\) × 24 = 12
⇒ b = \(\frac {1}{2}\)(5 – 19) = \(\frac {1}{2}\) × (-14) = -7
Take a – b = -19
a + b = 5, a – b = -19
⇒ a = \(\frac {1}{2}\)(5 + (-19) = \(\frac {1}{2}\) × (-14) = -7
⇒ b = \(\frac {1}{2}\)(5 – (-19)) = \(\frac {1}{2}\) × (5 + 19) = 12
∴ x² + 5x – 84 = (x + 12)(x – 7)

Question 6.
If each side of a square is decreased by 1 metre, its area becomes 49 square metres. What is the side of the original square?
Answer:
Let the side of the original square be x.
(x – 2)² = 49
⇒ x² – 4x + 4 = 49
⇒ x² – 4x – 45 = 0
a = 1, b = -4, c = -45

x is the length of the side of the square, so it cannot be negative.
∴ x = 9

Question 7.
The product of the first term and the third term of an arithmetic sequence with a common difference of 1 is 143. Find the first three terms of the sequence.
Answer:
If first term be x, then third term = x + 2
x(x + 2) = 143
⇒ x² + 2x = 143
⇒ x² + 2x – 143 = 0
a = 1, b = 2, c = -143
x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

If the first term x = 11, then the first three terms of the sequence are 11, 12, 13.
If the first term x = -13, then the first three terms of the sequence are -13, -12, -11.

Question 8.
1 added to the product of two consecutive even numbers gives 289. What are the numbers?
Answer:
If one number is x, then the next even number is x + 2.
x(x + 2) + 1 = 289
⇒ x² + 2x + 1 = 289
⇒ x² + 2x + 1 – 289 = 0
⇒ x² + 2x – 288 = 0
a = 1, b = 2, c = -288

If x = 16, then next number x + 2 = 18
If x = -18, then next number x + 2 = -16

Question 9.
The length of the rectangular sheet shown in the figure is 13 centimetres. From this sheet, cut off two square sheets of maximum size. The area of the remaining sheet is 15 square metres.

(a) If the breadth of the sheet is x, then what is the length of the remaining sheet?
(b) Form a second-degree equation and find the length and breadth of the remaining sheet.
Answer:
(a) Length of the remaining sheet = 13 – 2x
(b) Area of the remaining sheet = 15 square metres
x(13 – 2x) = 15
⇒ 13x – 2x² = 15
⇒ 2x² – 13x + 15 = 0
a = 2, b = -13, c = 15

If the breadth of the sheet is x = 5,
then length = 13 – 2x = 13 – 2 × 5 = 3
If breadth of the sheet is x = \(\frac {3}{2}\),
then length = 13 – 2x = 13 – 2 × \(\frac {3}{2}\) = 10

Question 10.
The sum of n natural numbers starting from 1 is \(\frac{n(n+1)}{2}\). Then, the sum of the consecutive natural numbers from 1 up to what number is 325?
Answer:
\(\frac{n(n+1)}{2}\) = 325
⇒ \(\frac{n^2+n}{2}\) = 325
⇒ n² + n = 650
⇒ n² + n – 650 = 0
n² + n – 650 = (n + a)(n + b) = n² + (a + b)n + ab
⇒ a + b = 1, ab = -650
(a – b)² = (a + b)² – 4ab
⇒ (a – b)² = 1² – 4 × (-650)
⇒ (a – b)² = 1 + 2600 = 2601
⇒ a – b = ±51
Take a – b = 51,
a + b = 1, a – b = 51
⇒ a = \(\frac {1}{2}\)(1 + 51) = \(\frac {1}{2}\) × 52 = 26
⇒ b = \(\frac {1}{2}\)(1 – 51) = \(\frac {1}{2}\) × (-50) = -25
Take a – b = -51,
a + b = 1, a – b = -51
⇒ a = \(\frac {1}{2}\)(1 – 51) = \(\frac {1}{2}\) × (-50) = -25
⇒ b = \(\frac {1}{2}\)(1 – (-51)) = \(\frac {1}{2}\) × (1 + 51) = 26
Using this, n² + n – 650 = (n + 26)(n – 25)
n² + n – 650 = 0
⇒ (n + 26)(n – 25) = 0
⇒ n = -26 or n = 25
n is the number of terms, so it cannot be negative.
∴ n = 25
The sum of the consecutive natural numbers from 1 up to 25 is 325.

Question 11.
In the figure, chord AB and CD are extended to meet at P. PB = 14 centimetres, AB = 5 centimetres, and CD = 15 centimetres. Find the length of PC.

Answer:
Let the length of PC be x,
PD = x + 15
PB = 14,
AB = 5
⇒ PA = 14 – 5 = 9
PA × PB = PC × PD
⇒ 9 × 14 = x(x + 15)
⇒ x² + 15x = 126
⇒ x² + 15x – 126 = 0
⇒ a = 1, b = 15, c = -126

Since x is the length of PC, and hence not negative.
So, x = 6
∴ The length of PC = 6 centimetres.

Question 12.
The difference between the perpendicular sides of a right triangle is 10 centimetres, and its area is 72 square centimetres. Then find the length of the perpendicular sides of the triangle.
Answer:
Let the length of one perpendicular side be x,
Then the other side is x + 10
Area = \(\frac {1}{2}\) × x(x + 10) = 72
⇒ x(x + 10) = 144
⇒ x² + 10x = 144
⇒ x² + 10x – 144 = 0
⇒ a = 1, b = 10, c = -144

Since x is the length of a side, it cannot be negative.
∴ x = 8
So, length of perpendicular sides = 8, 18

Question 13.
A rectangle is to be made by bending a 70-centimeter-long rod. The length of the diagonal is 25 centimetres. Then, find the length and breadth of the rectangle.
Answer:
Perimeter of the rectangle = 70 centimetres
Let the length of the rectangle be x,
⇒ 2(length + breadth) = 72
⇒ length + breadth = 36
⇒ x + breadth = 36
⇒ breadth = 36 – x
If the length of the diagonal is 25 centimetres,
x² + (36 – x)² = 25²
⇒ x² + 36² – 72x + x² = 252
⇒ 2x² – 72x + 1296 = 625
⇒ 2x² – 72x + 671 = 0
a = 2, b = -72, c = 671
Hence,

A rectangle diagonal of 25 cm cannot be made.

Question 14.
The terms of an arithmetic sequence with a common difference of 4 are positive numbers. The product of two consecutive terms of the sequence is equal to its sum.
(a) If one term is x, then what is the next term?
(b) Find the terms.
Answer:
(a) Common difference = 4
If one term is x, next term = x + 4
(b) x(x + 4) = x + x + 4
⇒ x² + 4x = 2x + 4
⇒ x² + 4x – 2x – 4 = 0
⇒ x² + 2x – 4 = 0
a = 1, b = 2, c = -4
Hence,

If x = -1 + √5, then next term = x + 4
= -1 + √5 + 4
= 3 + √5
Sequence: -1 + √5, 3 + √5, 7 + √5,…
If x = -1 – √5, then next term = x + 4
= -1 – √5 + 4
= 3 – √5
Sequence: -1 – √5, 3 – √5, 7 – √5,….

Question 15.
When 16 is added to the sum of several terms of the arithmetic sequence 9, 11, 13…, we get 256. How many terms are added?
Answer:
When 16 is added to the sum of several terms of the arithmetic sequence 9, 11, 13,…, we get 256.
That is, the sum of terms of the arithmetic sequence = 256 – 16 = 240
First term (a) = 9
Common difference (d) = 2
x_n = dn + (f – d)
= 2n + (9 – 2)
= 2n + 7
If the algebraic form of an arithmetic sequence is an + b,
then the sum of first n terms of the sequence = a\(\frac {n}{2}\)(n + 1) + nb
= 2\(\frac {n}{2}\)(n + 1) + n × 7
= n(n + 1) + 7n
= n² + n + 7n
= n² + 8n
⇒ n² + 8n = 240
⇒ n² + 8n – 240 = 0
a = 1, b = 8, c = -240

n is the number of terms, so it cannot be negative.
Thus, n = 12
12 terms are added.