If a = 5 + 2 5 - 2 and b = 5 - 2 5 + 2 show that 3 a 2 + 4 a b - 3 b 2 = 4 + 56 3 10

Given: a = 5 + 2 5 - 2 Since denominator is an irrational number we have to rationalize it by multiplying its numerator and denominator by its rationalizing factor we get So, a = 5 + 2 5 - 2 × 5 + 2 5 + 2 [According to the formula a 2 - b 2 = ( a + b ) ( a - b ) ] = ( 5 + 2 ) 2 5 2 - 2 2 = 5 + 2 10 + 2 5 - 2 = 7 + 2 10 3 . Also, given that, b = 5 - 2 5 + 2 Since denominator is an irrational number we have to rationalize it by multiplying its numerator and denominator by its rationalizing factor we get b = 5 - 2 5 + 2 × 5 - 2 5 - 2 [According to the formula a 2 - b 2 = ( a + b ) ( a - b ) ] = ( 5 - 2 ) 2 5 2 - 2 2 = 5 - 2 10 + 2 5 - 2 = 7 - 2 10 3 . a 2 = 7 + 2 10 3 2 = 49 + 28 10 + 40 9 = 89 + 28 10 9 . b 2 = 7 - 2 10 3 2 = 49 - 28 10 + 40 9 = 89 - 28 10 9 . Now, L H S = 3 a 2 + 4 a b - 3 b 2 = 3 × 89 + 28 10 9 + 4 × 5 + 2 5 - 2 × 5 - 2 5 + 2 - 3 × 89 - 28 10 9 = 89 + 28 10 3 + 4 - 89 - 28 10 3 = 4 + 89 + 28 10 - 89 + 28 10 3 = 4 + 56 3 10 = RHS . (proved)

If 3 4 + 3 7 7 = x , 3 2 + 3 5 7 = y and z = x + y , then 4 z 2 + 3 =

Given: 3 4 + 3 7 7 = x , 3 2 + 3 5 7 = y To find: 4 z 2 + 3 We have, z = x + y By squaring on both the sides, ⇒ z 2 = x + y = 3 4 + 3 7 7 + 3 2 + 3 5 7 = 3 4 + 3 7 7 + 3 2 + 3 5 7 = 81 + 2187 + 9 + 243 7 = 2268 + 252 7 = 324 × 7 + 36 × 7 7 = 18 2 × 7 + 6 2 × 7 7 = 18 7 + 6 7 7 = 7 18 + 6 7 = 24 Now, 4 z 2 + 3 = 4 24 + 3 = 4 27 = 108 Therefore, 4 z 2 + 3 = 108

If x = a 2 + b 2 + a 2 - b 2 a 2 + b 2 - a 2 - b 2 , then show that b 2 x 2 - 2 a 2 x + b 2 = 0 .

Given, x = a 2 + b 2 + a 2 - b 2 a 2 + b 2 - a 2 - b 2 . To show: b 2 x 2 - 2 a 2 x + b 2 = 0 Now, rationalizing x x = a 2 + b 2 + a 2 - b 2 a 2 + b 2 - a 2 - b 2 × a 2 + b 2 + a 2 - b 2 a 2 + b 2 + a 2 - b 2 = a 2 + b 2 + a 2 - b 2 2 a 2 + b 2 2 - a 2 - b 2 2 = a 2 + b 2 2 + a 2 - b 2 2 + 2 a 2 + b 2 a 2 - b 2 a 2 + b 2 - a 2 + b 2 = a 2 + b 2 + a 2 - b 2 + 2 a 4 - b 4 2 b 2 = a 2 + a 4 - b 4 b 2 ⇒ x = a 2 + a 4 - b 4 b 2 Substituting this value of x in the given equation, we get b 2 a 2 + a 4 - b 4 b 2 2 - 2 a 2 a 2 + a 4 - b 4 b 2 + b 2 = a 2 + a 4 - b 4 2 b 2 - 2 a 4 + 2 a 2 a 4 - b 4 b 2 + b 2 = a 4 + a 4 - b 4 + 2 a 2 a 4 - b 4 - 2 a 4 - 2 a 2 a 4 - b 4 + b 4 b 2 = 0 Hence, proved.

Complete the table: Q - Rational numbers I R - Irrational numbers. Type of Number Number Sum(Q or IR) Difference(Q or IR) Product(Q or IR) Quotient(Q or IR) 1 + 2 , 1 - 2 3 , 3 + 2

Let us consider the numbers as 1 + 2 and 1 - 2 Sum = 1 + 2 + 1 - 2 = 2 We know that rational numbers are numbers that can be written in the form of p q , where p and q are integers, q ≠ 0 So, the sum is a rational number. Difference = 1 + 2 - 1 + 2 = 2 2 Irrational numbers are numbers that can be written in the form of p q , where p and q are integers, q ≠ 0 We know that 2 is an irrational number, so 2 2 is an irrational number. Product = 1 + 2 1 - 2 = 1 - 2 2 = - 1 which is a rational number. So, the product is a rational number. Quotient = 1 + 2 1 - 2 Rationalise the denominator, we get Q = 1 + 2 1 - 2 × 1 + 2 1 + 2 = - 1 + 2 + 2 2 = - 3 + 2 2 Since - 3 + 2 2 is an irrational number, so the quotient is an irrational number. Consider the number 3 and 3 + 2 Sum = 3 + 3 + 2 = 6 + 2 which is an irrational number So, sum of the given two numbers are irrational. Difference = 3 - 3 + 2 = - 2 which is an irrational number. So, the difference of the given two numbers are irrational. Product = 3 3 + 2 = 9 + 3 2 which is an irrational number So, the product of the given two numbers are irrational number Quotient = 3 3 + 2 Rationalise the denominator, we get = 3 3 - 2 3 + 2 3 - 2 = 9 - 3 2 7 Since 9 - 3 2 is an irrational number, 9 - 3 2 7 is an irrational number. Hence, 3 3 + 2 is an irrational number. Type of Number Number Sum(Q or IR) Difference(Q or IR) Product(Q or IR) Quotient(Q or IR) 1 + 2 , 1 - 2 2 (Q) 2 2 (IR) - 1 (Q) - 3 + 2 2 (IR) 3 , 3 + 2 6 + 2 (IR) - 2 (IR) 9 + 3 2 (IR) 9 - 3 2 7 (IR)

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Mathematics Textbook of Competency Based Questions for Class IX>Chapter 1 - Number Systems>Exercise>Q 2

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Which of the following statements is true?

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Important Questions on Number Systems

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Mathematics Textbook of Competency Based Questions for Class IX>Chapter 1 - Number Systems>Exercise>Q 7

Find the value of

\sqrt{15}

by geometric method.

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Mathematics Textbook of Competency Based Questions for Class IX>Chapter 1 - Number Systems>Exercise>Q 8

Two persons started from a common point $(0, 0)$ in the figure below, with the same speed. One person $(X)$ went along a straight line path to $P$ , which is $25 m$ East and $32 km$ North. The second person $(Y)$ went due East.

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To which of these positions person $Y$ will be the closest when person $(X)$ is at $P$ ?

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Mathematics Textbook of Competency Based Questions for Class IX>Chapter 1 - Number Systems>Exercise>Q 12

If $\frac{\sqrt{7} + 2 \sqrt{3}}{2 \sqrt{7} - \sqrt{5}} = \frac{c + \sqrt{p} + \sqrt{q} + \sqrt{r}}{23} (p < q < r)$ where $p, q, r$ are rational numbers, then $q + r - p =$