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If x = and y = ; Find: (i) x2 (ii) y2 (iii) xy (iv) x2 + y2 = xy Solution: (i) (ii) (iii) We know that (iv) x2 + y2 = xy By substituting the values = 161 – 72√5 + 161 + 72√5 + 1 So, we get = 322 + 1 = 323 7. Simplify: Solution: We have, It can be written as So, we get 2. If (2 + √5)/(2 – √5) = x and (2 – √5)/(2 + √5) = y; find the value of x2 – y2. Solution: We have, Using the formula a2 – b2 = (a + b) (a – b) So, we get Similarly, Using the formula a2 – b2 = (a + b) (a – b) By further calculation Here, x2 – y2 = (-9 – 4√5)2 – (-9 + 4√5)2 Expanding using the formula, we get = 81 + 72√5 + 80 – (81 – 72√5 + 80) = 81 + 72√5 + 80 – 81 + 72√5 – 80 = 144√5 Exercise 1D PAGE: 22 1. If x = 2√3 + 2√2, find: (i) 1/x (ii) x + 1/x (iii) (x + 1/x)2 Solution: (i) (ii) (iii) 9. Automaticallygenerate support based on overhang angles, then add or removesupports as needed. – Customize the placement, size, and angle of supportmaterial. – Snap off supports easily without damaging your part. – Enjoy the best of both worlds — sufficient support and cleanersurface finishes. Optimized Dual Extrusion Create stunning dual-color parts with Simplify3D’s Dual ExtrusionWizard. Rationalize the denominators of: Solution: (i) (3/√5) x (√5/√5) = 3√5/5 (ii) (2√3/√5) x (√5/√5) = 2√15/5 (iii) (iv) (v) (vi) (vii) = 5 – 2√6 (viii) (ix) 4. Write a pair of irrational numbers whose difference is irrational. Solution: √3 + 2 and √2 – 3 are irrational numbers whose sum is irrational. Here, Difference = (√3 + 2) – (√2 – 3) = √3 – √2 + 2 + 3 = √3 – √2 + 5 Hence, the resultant is irrational. 10. (1) Take a = 3c, x = 5z By squaring on both sides a2 = 9c2, x2 = 25z2 Using equation (a) 3b2 = 9c2, 5y2 = 25z2 By further simplification b2 = 3c2, y2 = 5z2 Here, B2 and y2 are odd as 3c2 and 5z2 are odd. b and y are odd …… (2) Using equation (1) and (2) we know that a, b, x, y are odd integers. a, b and x, y have common factors 3 and 5 which contradicts our assumption that a/b and x/y are rational a, b and x, y do not have any common factors a/b and x/y is not rational √3 and √5 are irrational. 6. Insert two irrational numbers between 5 and 6. Solution: Let’s write 5 and 6 as square root Then, 5 = √25 and 6 = √36 Now, take the numbers √25 < √26 < √27 < √28 < √29 < √30 < √31 < √32 < √33 < √34 < √35 < √36 Hence, any two irrational numbers between 5 and 6 is √29 and √30 17. simplify 3 3 2 download free. If x = 5 − 2√6, find: x2 + 1/x2 Solution: It is given that x = 5 − 2√6 We should find the value of (x2 + 1/x2) So, x = 5 − 2√6, we get Using the formula (a – b) (a + b) = a2 – b2 Here, (x – 1/x) = (5 – 2√6) – (5 + 2√6) = 5 – 2√6 – 5 – 2√6 = -4√6 … (2) Now, Consider (x – 1/x)2 Using the equation (a – b)2 = a2 + b2 – 2ab (x – 1/x)2 = x2 + 1/x2 – 2(x)(1/x) (x – 1/x)2 = x2 + 1/x2 – 2 (x – 1/x)2 + 2 = x2 + 1/x2 … (3) From equations (2) and (3), we get x2 + 1/x2 = (-4√6)2 + 2 = 96 + 2 = 98 11. These Selina solutions for Class 9 Maths help the students in understanding all the concepts in a better way. (a) Here, a2 and x2 are odd as 3b2 and 5y2 are odd. a and x are odd …. Class 9 is an important phase of a student’s life, the concepts which are taught in Class 9 are vital to be understood as these concepts are continued in Class 10. Show that: Solution: We have, 10. Can it be written in the form p/q, where p and q are integers and q ≠ 0? Solution: Yes, zero is a rational number. It can be written in the form of 𝑝/q, where p and q are integers and q ≠ 0 ⇒ 0 = 0/1. 2. Write a pair of irrational numbers whose product is irrational. Solution: Let us take two irrational numbers (5 + √2) and (√5 – 2) Here the product = (5 + √2) × (√5 – 2) By further calculation = 5 √5 – 10 + √10 – 2√2 which is irrational. 12. Evaluate, correct to one place of decimal. Given universal set is {-6, -5¾, -√4, -3/5, -3/8, 0, 4/5, 1, 1⅔, √8, 3.01, π, 8.47} From the given set, find: (i) Set of Rational numbers (ii) Set of irrational numbers (iii) Set of integers (iv) Set of non-negative integers Solution: (i) First find the set of rational numbers Rational numbers are numbers of the form p/q, where q ≠ 0 U = {-6, -5¾, -√4, -3/5, -3/8, 0, 4/5, 1, 1⅔, √8, 3.01, π, 8.47} Here, -5¾, -3/5, -3/8, 4/5 and 1⅔ are of the from p/q Therefore, they are rational numbers The set of integers is a subset of rational numbers, -6, 0 and 1 are also rational numbers Here, decimal numbers 3.01 and 8.47 are also rational numbers as they are terminating decimals Also, -√4 = -2 as square root of 4 is 2 Thus, -2 belongs to the set of integers From the above set, the set of rational numbers is Q, Q = {-6, -5¾, -√4, -3/5, -3/8, 0, 4/5, 1, 1⅔, 3.01, 8.47} (ii) First find the set of irrational numbers Irrational numbers are numbers which are not rational From the above subpart, we know that the set of rational numbers is Q, Q = {-6, -5¾, -√4, -3/5, -3/8, 0, 4/5, 1, 1⅔, 3.01, 8.47} Here the set of irrational numbers is the set of complement of the rational numbers over real numbers The set of irrational numbers is U – Q = {√8, π} (iii) First find the set of integers Set of integers consists of zero, the natural numbers and their additive inverses Set of integers is Z Z = {…, -3, -2, -1, 0, 1, 2, 3, …} Here, the set of integers is U ⋂ Z = {-6, -√4, 0, 1} (iv) First find the set of non-negative integers Set of non-negative integers consists of zero and the natural numbers Set of non-negative integers is Z+ and Z+ = {0, 1, 2, 3, …} Set of integers is U ⋂ Z+ = {0, 1} 5. Read and learn the Chapter 1 of Selina textbook to learn more about Rational And Irrational Numbers along with the concepts covered in it. Without doing any actual division, find which of the following rational numbers have terminating decimal representation: (i) 7/16 (ii) 23/125 (iii) 9/14 (iv) 32/45 (v) 43/50 (vi) 17/40 (vii) 61/75 (viii) 123/250 Solution: (i) Given number is 7/16 16 = 2 x 2 x 2 x 2 = 24 = 24 x 50 So, 16 can be expressed as 2m x 5n Hence, 7/16 is convertible into the terminating decimal (ii) Given number is 23/125 125 = 5 x 5 x 5 = 53 = 20 x 53 So, 125 can be expressed as 2m x 5n Hence, 23/125 is convertible into the terminating decimal (iii) Given number is 9/14 14 = 2 x 7 = 21 x 71 So, 14 cannot be expressed as 2m x 5n Hence, 9/14 is not convertible into the terminating decimal (iv) Given number is 32/45 45 = 3 x 3 x 5 = 32 x 51 So, 45 cannot be expressed as 2m x 5n Hence, 32/45 is not convertible into the terminating decimal (v) Given number is 43/50 50 = 2 x 5 x 5 = 21 x 52 So, 50 can be expressed as 2m x 5n Hence, 43/50 is convertible into the terminating decimal (vi) Given number is 17/40 40 = 2 x 2 x 2 x 5 = 23 x 51 So, 40 can be expressed as 2m x 5n Hence, 17/40 is convertible into the terminating decimal (vii) Given number is 61/75 75 = 3 x 5 x 5 = 31 x 52 So, 75 cannot be expressed as 2m x 5n Hence, 61/75 is not convertible into the terminating decimal (viii) Given number is 123/250 250 = 2 x 5 x 5 x 5 = 21 x 53 So, 250 can be expressed as 2m x 5n Hence, 123/250 is convertible into the terminating decimal Exercise 1(B) PAGE: 13 1. ∛40 is not a surd (vi) ∛-125 = ∛(-5 x -5 x -5) = -5 It is rational Therefore, ∛-125 is not a surd (vii) π is irrational. Therefore, √π is not a surd. (viii) 3 + √2 is irrational 2. Download pdf of Class 9 Maths Chapter 1 Selina Solutions from the given links. Download PDF of Selina Solutions for Class 9 Maths Chapter 1:-Download Here Exercise 1(A) PAGE: 4 1. simplify 3 3 2 download free. Insert five irrational numbers between 2√5 and 3√3. Solution: Here, 2√5 = √(22 x 5) = √(4 x 5) = √20 and 3√3 = √(32 x 3) = √(9 x 3) = √27 Now, take the numbers √20 < √21 < √22 < √23 < √24 < √25 < √26 < √27 Hence, any five irrational numbers between 2√5 and 3√3 are: √21, √22, √23, √24 and √26 18. Use method of contradiction to show that √𝟑 and √𝟓 are irrational. Solution: Consider √3 and √5 as rational numbers √3 = a/b and √5 = x/y (where a, b, x, y ∈ Z and b, y ≠ 0) By squaring on both sides, we have 3 = a2/b2 , 5 = x2/y2 3b2 = a2 , 5y2 = x2 …. Chapter 1 of Class 9 takes the students to the different sets of numbers, the Rational And Irrational Numbers.

simplify 3 3 2 download free

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