First half of Exercise 5(a)…

Question 4:

(i) If x + \frac{1}{x} = 6 , find (a) x^2 + \frac{1}{x^2}    (b)  x^4 + \frac{1}{x^4}

Answer:

(a) x + \frac{1}{x} = 6

\Rightarrow \Big( x + \frac{1}{x} \Big)^2 = 36

\Rightarrow x^2 + \frac{1}{x^2} + 2 = 36

\Rightarrow x^2 + \frac{1}{x^2} = 34

(b)  x^2 + \frac{1}{x^2} = 34

\Rightarrow \Big( x^2 + \frac{1}{x^2} \Big)^2 = 34^2

\Rightarrow x^4 + \frac{1}{x^4} + 2 = 1156

\Rightarrow x^4 + \frac{1}{x^4} = 1154

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(ii) If x = \frac{1}{x - 2\sqrt{3}} , find  (a) x- \frac{1}{x}     (b) x- \frac{1}{x}    (c) x^2- \frac{1}{x^2}

Answer:

(a)  x = \frac{1}{x - 2\sqrt{3}}

x(x - 2\sqrt{3}) = 1

x^2 - 1 = 2\sqrt{3}x

\frac{x^2-1}{x} = 2\sqrt{3}

x- \frac{1}{x} = 2\sqrt{3}

(b)  Since (a+b)^2 - (a-b)^2 = 4ab

Therefore \Big( x + \frac{1}{x} \Big)^2 - \Big( x - \frac{1}{x} \Big)^2 = 4

\Rightarrow\Big( x + \frac{1}{x} - \Big)^2 - (2\sqrt{3})^2 = 4

\Rightarrow\Big( x + \frac{1}{x} - \Big)^2  = 4 + (2\sqrt{3})^2

\Rightarrow\Big( x + \frac{1}{x} - \Big)^2  =16

\Rightarrow x + \frac{1}{x} - = \pm 4

(c)  x- \frac{1}{x} = 2\sqrt{3}

\Rightarrow x + \frac{1}{x} - = \pm 4

Therefore (x- \frac{1}{x} ) (x + \frac{1}{x} ) = 2\sqrt{3} \times \pm 4 = \pm 8\sqrt{3}

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(iii) If x^2+ \frac{1}{x^2} = 27 , find (a) x+ \frac{1}{x}    (b) x- \frac{1}{x}

Answer:

(a)  x^2+ \frac{1}{x^2} = 27

\Big( x + \frac{1}{x} \Big)^2 = x^2 + \frac{1}{x^2} +2

\Big( x + \frac{1}{x} \Big)^2 = 27 +2

\Big( x + \frac{1}{x} \Big) = \pm \sqrt{29}

(b)  x^2+ \frac{1}{x^2} = 27

\Big( x - \frac{1}{x} \Big)^2 = x^2 + \frac{1}{x^2} -2

\Big( x - \frac{1}{x} \Big)^2 = 27 - 2

\Big( x - \frac{1}{x} \Big) = \pm 5

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(iv) x + \frac{1}{x} = 11 , find the value of  x^2 + \frac{1}{x^2}

Answer:

x + \frac{1}{x} = 11

\Rightarrow \Big( x + \frac{1}{x} \Big)^2 = 121

\Rightarrow x^2 + \frac{1}{x^2} + 2 = 121

\Rightarrow x^2 + \frac{1}{x^2} = 119

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(v) x - \frac{1}{x} = -1 , find the value of  x^2 + \frac{1}{x^2}

Answer:

x - \frac{1}{x} = -1

\Rightarrow \Big( x - \frac{1}{x} \Big)^2 = 1

\Rightarrow x^2 + \frac{1}{x^2} - 2 = 1

\Rightarrow x^2 + \frac{1}{x^2} = 3

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(vi) x + \frac{1}{x} = \sqrt{5} , find the value of  x^2 + \frac{1}{x^2} and x^4 + \frac{1}{x^4}

Answer:

x + \frac{1}{x} = \sqrt{5}

\Rightarrow \Big( x + \frac{1}{x} \Big)^2 = 5

\Rightarrow x^2 + \frac{1}{x^2} + 2 = 5

\Rightarrow x^2 + \frac{1}{x^2} = 3

Now

x^2 + \frac{1}{x^2} = 3

\Rightarrow \Big( x^2 + \frac{1}{x^2} \Big)^2 = 9

\Rightarrow x^4 + \frac{1}{x^4} + 2 = 9

\Rightarrow x^2 + \frac{1}{x^2} = 7

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(vii) If x^2 + \frac{1}{x^2} = 66 , find the value of x- \frac{1}{x}

Answer:

x^2+ \frac{1}{x^2} = 66

\Big( x - \frac{1}{x} \Big)^2 = x^2 + \frac{1}{x^2} -2

\Big( x - \frac{1}{x} \Big)^2 = 66 - 2

\Big( x - \frac{1}{x} \Big) = \pm 8

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(viii) If x^2 + \frac{1}{x^2} = 79 , find the value of x+ \frac{1}{x}

Answer:

x^2+ \frac{1}{x^2} = 79

\Big( x + \frac{1}{x} \Big)^2 = x^2 + \frac{1}{x^2} +2

\Big( x + \frac{1}{x} \Big)^2 = 79 +2

\Big( x + \frac{1}{x} \Big) = \pm 9

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(ix) If a^2-3a-1=0 , find the value of a^2 + \frac{1}{a^2}

Answer:

a^2-3a-1=0

\Rightarrow a^2 -1 = 3a

\Rightarrow \frac{a^2-1}{a} = 3

\Rightarrow a - \frac{1}{a} = 3

\Big( a - \frac{1}{a} \Big)^2 = a^2 + \frac{1}{a^2} -2

\Big( x - \frac{1}{x} \Big)^2 = 9 - 2

\Big( x - \frac{1}{x} \Big) = \pm 7

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(x) If x^2 + \frac{1}{x^2} = 7 , find the value of 3x^2- \frac{3}{x^2}

Answer:

x^2+ \frac{1}{x^2} = 7

\Big( x - \frac{1}{x} \Big)^2 = x^2 + \frac{1}{x^2} -2

\Big( x - \frac{1}{x} \Big)^2 = 7 - 2

\Big( x - \frac{1}{x} \Big) = \pm \sqrt{5}

Now 3x^2- \frac{3}{x^2} = 3 ( x^2- \frac{1}{x^2} )  = \pm 3\sqrt{5}

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(xi) If a = \frac{1}{a-5} , find the value of (a) a - \frac{1}{a}    (b) a + \frac{1}{a}   (c) a^2 + \frac{1}{a^2}

Answer:

(a)  a = \frac{1}{a-5}

a^2 - 5a = 1

\Rightarrow a^2 - 1 = 5a

\Rightarrow a - \frac{1}{a} = 5

(b) \Rightarrow \Big( a + \frac{1}{a} \Big)^2 =  \Rightarrow \Big( a - \frac{1}{a} \Big)^2 +4

\Rightarrow \Big( a + \frac{1}{a} \Big)^2 = 29

\Rightarrow \Big( a + \frac{1}{a} \Big) = \pm \sqrt{29}

(c) a + \frac{1}{a} = \pm \sqrt{29}

\Rightarrow \Big( a + \frac{1}{a} \Big)^2 = 29

\Rightarrow a^2 + \frac{1}{a^2} + 2 = 29

\Rightarrow a^2 + \frac{1}{a^2} = 27

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(xii) If a^2-3a+1 = 0 , find the values of (a) a^2 + \frac{1}{a^2}    (b) a^3 + \frac{1}{a^3}

Answer:

a^2-3a+1 = 0

a^2+1 = 3a

\Rightarrow a^2 + 1 = 3a

\Rightarrow a + \frac{1}{a} = 3

(a)  a + \frac{1}{a} = 3

\Rightarrow \Big( a + \frac{1}{a} \Big)^2 = 9

\Rightarrow a^2 + \frac{1}{a^2} + 2 = 9

\Rightarrow a^2 + \frac{1}{a^2} = 7

(b)  a + \frac{1}{a} = 3

\Rightarrow \Big( a + \frac{1}{a} \Big)^3 = a^3 + \frac{1}{a^3} + 3. a . \frac{1}{a} (a + \frac{1}{a} )

\Rightarrow \Big( a + \frac{1}{a} \Big)^3 = a^3 + \frac{1}{a^3} + 3 (a + \frac{1}{a} )

a^3 + \frac{1}{a^3} = 3^3 -3(3) = 27 -9 = 18

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(xiii) If x = 5 - 2\sqrt{6} find the value of  \sqrt{x}+ \frac{1}{\sqrt{x}}

Answer:

x = 5 - 2\sqrt{6}

\Big( \sqrt{x} + \frac{1}{\sqrt{x}} \Big)^2 = x + \frac{1}{x} + 2

= 5 - 2\sqrt{6} + \frac{1}{5 - 2\sqrt{6}} + 2

= \frac{25+24-20\sqrt{6}+1+10-4\sqrt{6}}{5 - 2\sqrt{6}}

= \frac{60-24\sqrt{6}}{5 - 2\sqrt{6}}

= \frac{12(5 - 2\sqrt{6})}{5 - 2\sqrt{6}} = 12

Therefore \sqrt{x}+ \frac{1}{\sqrt{x}} = \pm \sqrt{12}

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(xiv) If a+b+c = 0 and a^2+b^2+c^2=16 , find the value of ab+bc+ca

Answer:

Given: a+b+c = 0 and a^2+b^2+c^2=16

We know: (a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)

\Rightarrow ab+bc+ca = \frac{0-16}{2} = -8

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(xv) If a^2+b^2+c^2=16 and ab+bc+ca=10 , find the value of a+b+c

Answer:

Given: a^2+b^2+c^2=16 and ab+bc+ca=10

We know: (a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)

\Rightarrow (a+b+c)^2 = 16+2 \times 10 = 36

\Rightarrow a+b+c = \pm 6

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(xvi) If a+b+c = 9 and ab+bc+ca=23 , find the value of a^2+b^2+c^2

Answer:

Given: a+b+c = 9 and ab+bc+ca=23

We know: (a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)

\Rightarrow a^2 + b^2 + c^2 = 81-46 = 35

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(xvii) If x+y-z=4 and x^2+y^2+z^2=38 , find the value of xy-yz-zx

Answer:

Given: x+y-z=4 and x^2+y^2+z^2=38

We know: (x+y-z)^2 = x^2 + y^2 + z^2 + 2(xy - yz - zx)

\Rightarrow xy - yz - zx = \frac{16-38}{2} = -11

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(xviii) If x^2 + \frac{1}{x^2} = 7 , find the value of x^3+ \frac{1}{x^3}

Answer:

Given x^2 + \frac{1}{x^2}  = 7

\Rightarrow \Big( x + \frac{1}{x} \Big)^2 =x^2 + \frac{1}{x^2} + 2

\Rightarrow \Big( x + \frac{1}{x} \Big)^2 =9

\Rightarrow \Big( x + \frac{1}{x} \Big) =  \pm 3

If  x + \frac{1}{x} = 3

\Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3. x . \frac{1}{x} (x + \frac{1}{x} )

\Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3 (x + \frac{1}{x} )

x^3 + \frac{1}{x^3} = 3^3 -3(3) = 18

Similarly, if  x + \frac{1}{x} = -3

Then x^3 + \frac{1}{x^3} = (-3)^3 -3(-3) = -18

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(xix) If x^2 + \frac{1}{x^2} = 83 , find the value of x^3- \frac{1}{x^3}

Answer:

Given x^2 + \frac{1}{x^2}  = 83

\Rightarrow \Big( x - \frac{1}{x} \Big)^2 =x^2 + \frac{1}{x^2} - 2

\Rightarrow \Big( x - \frac{1}{x} \Big)^2 =81

\Rightarrow \Big( x - \frac{1}{x} \Big) =  \pm 9

If  x - \frac{1}{x} = 9

\Rightarrow \Big( x - \frac{1}{x} \Big)^3 = x^3 - \frac{1}{x^3} - 3. x . \frac{1}{x} (x - \frac{1}{x} )

\Rightarrow \Big( x - \frac{1}{x} \Big)^3 = x^3 - \frac{1}{x^3} - 3 (x - \frac{1}{x} )

x^3 - \frac{1}{x^3} = 9^3 +3(9) = 756

Similarly, if  x - \frac{1}{x} = -9

Then x^3 - \frac{1}{x^3} = (-9)^3 +3(-9) = -756

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(xx) If x^4 + \frac{1}{x^4} = 47 , find the value of x^3+ \frac{1}{x^3}

Answer:

Given x^4 + \frac{1}{x^4}  = 47

\Rightarrow \Big( x^2 + \frac{1}{x^2} \Big)^2 =x^4 + \frac{1}{x^4} + 2

\Rightarrow \Big( x^2 + \frac{1}{x^2} \Big)^2 =49

\Rightarrow \Big( x^2 + \frac{1}{x^2} \Big) =  7

Now \Big( x + \frac{1}{x} \Big)^2 =x^2 + \frac{1}{x^2} + 2

\Rightarrow \Big( x + \frac{1}{x} \Big)^2 =9

\Rightarrow \Big( x + \frac{1}{x} \Big) =  \pm 3

If  x + \frac{1}{x} = 3

\Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3. x . \frac{1}{x} (x + \frac{1}{x} )

\Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3 (x + \frac{1}{x} )

x^3 + \frac{1}{x^3} = 3^3 -3(3) = 18

Similarly, if  x + \frac{1}{x} = -3

Then x^3 + \frac{1}{x^3} = (-3)^3 -3(-3) = -18

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(xxi) If a+b = 10 and ab = 21 , find the value of a^3+b^3

Answer:

Given a+b = 10 and ab = 21

(a+b)^3 = a^3 + b^3 +3ab(a+b)

\Rightarrow a^3 + b^3 = 10^3 - 3 \times 21 \times 10 = 1000-630 = 370

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(xxii) If a-b = 4 and ab = 21 , find the value of a^3-b^3

Answer:

Given a-b = 4 and ab = 21

(a-b)^3 = a^3 - b^3 - 3ab(a-b)

\Rightarrow a^3 - b^3 = 4^3 + 3 \times 21 \times 4 = 64+252 = 316

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(xxiii) x + \frac{1}{x} = 5 , find x^3 + \frac{1}{x^3}

Answer:

If  x + \frac{1}{x} = 5

\Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3. x . \frac{1}{x} (x + \frac{1}{x} )

\Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3 (x + \frac{1}{x} )

x^3 + \frac{1}{x^3} = 5^3 -3(5) = 110

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(xxiv) x - \frac{1}{x} = 7 , find x^3 - \frac{1}{x^3}

Answer:

If  x - \frac{1}{x} = 7

\Rightarrow \Big( x - \frac{1}{x} \Big)^3 = x^3 - \frac{1}{x^3} - 3. x . \frac{1}{x} (x - \frac{1}{x} )

\Rightarrow \Big( x - \frac{1}{x} \Big)^3 = x^3 - \frac{1}{x^3} - 3 (x - \frac{1}{x} )

x^3 - \frac{1}{x^3} = 7^3 + 3(7) = 364

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(xxv)  x^2 + \frac{1}{x^2} = 51 , find x^3 - \frac{1}{x^3}

Answer:

Given x^2 + \frac{1}{x^2}  = 51

\Rightarrow \Big( x - \frac{1}{x} \Big)^2 =x^2 + \frac{1}{x^2} - 2

\Rightarrow \Big( x - \frac{1}{x} \Big)^2 =49

\Rightarrow \Big( x - \frac{1}{x} \Big) =  \pm 7

If  x - \frac{1}{x} = 7

\Rightarrow \Big( x - \frac{1}{x} \Big)^3 = x^3 - \frac{1}{x^3} - 3. x . \frac{1}{x} (x - \frac{1}{x} )

\Rightarrow \Big( x - \frac{1}{x} \Big)^3 = x^3 - \frac{1}{x^3} - 3 (x - \frac{1}{x} )

x^3 - \frac{1}{x^3} = 7^3 + 3(7) = 364

Similarly, if  x - \frac{1}{x} = -7

Then x^3 - \frac{1}{x^3} = (-7)^3 +3(-7) = -364

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(xxvi) x^2 + \frac{1}{x^2} = 98 , find x^3 + \frac{1}{x^3}

Answer:

Given x^2 + \frac{1}{x^2} = 98

\Rightarrow \Big( x + \frac{1}{x} \Big)^2 =x^2 + \frac{1}{x^2} + 2

\Rightarrow \Big( x + \frac{1}{x} \Big)^2 =100

\Rightarrow \Big( x + \frac{1}{x} \Big) = 10

Now x + \frac{1}{x} = \pm 10

\Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3. x . \frac{1}{x} (x + \frac{1}{x} )

\Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3 (x + \frac{1}{x} )

x^3 + \frac{1}{x^3} = 10^3 -3(10) = 100-30 = 970

Similarly, if  x + \frac{1}{x} = -10

Then x^3 + \frac{1}{x^3} = (-10)^3 -3(-10) = -970

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(xxvii) x^4 + \frac{1}{x^4} = 119 , find x^3 - \frac{1}{x^3}

Answer:

Given x^4 + \frac{1}{x^4} = 119

\Rightarrow \Big( x^2 + \frac{1}{x^2} \Big)^2 =x^4 + \frac{1}{x^4} + 2

\Rightarrow \Big( x^2 + \frac{1}{x^2} \Big)^2 =121

\Rightarrow \Big( x^2 + \frac{1}{x^2} \Big) =  11

Given x^2 + \frac{1}{x^2} = 11

\Rightarrow \Big( x - \frac{1}{x} \Big)^2 =x^2 + \frac{1}{x^2} - 2

\Rightarrow \Big( x - \frac{1}{x} \Big)^2 =9

\Rightarrow \Big( x - \frac{1}{x} \Big) =  \pm 3

If  x - \frac{1}{x} = 3

\Rightarrow \Big( x - \frac{1}{x} \Big)^3 = x^3 - \frac{1}{x^3} - 3. x . \frac{1}{x} (x - \frac{1}{x} )

\Rightarrow \Big( x - \frac{1}{x} \Big)^3 = x^3 - \frac{1}{x^3} - 3 (x - \frac{1}{x} )

x^3 - \frac{1}{x^3} = 3^3 +3(3) = 36

Similarly, if  x - \frac{1}{x} = -3

Then x^3 - \frac{1}{x^3} = (-3)^3 +3(-3) = -36

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(xxviii) x + \frac{1}{x} = 5 , find x^2 + \frac{1}{x^2} x^3 + \frac{1}{x^3} and x^4 + \frac{1}{x^4}

Answer:

x + \frac{1}{x} = 5

\Rightarrow \Big( x + \frac{1}{x} \Big)^2 = 25

\Rightarrow x^2 + \frac{1}{x^2} + 2 = 25

\Rightarrow x^2 + \frac{1}{x^2} = 23

\Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3. x . \frac{1}{x} (x + \frac{1}{x} )

\Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3 (x + \frac{1}{x} )

x^3 + \frac{1}{x^3} = 5^3 -3(5) = 110

Now x^2 + \frac{1}{x^2} = 23

\Rightarrow \Big( x^2 + \frac{1}{x^2} \Big)^2 = 529

\Rightarrow x^4 + \frac{1}{x^4} + 2 = 529

\Rightarrow x^4 + \frac{1}{x^4} = 527

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(xxix) x^4 + \frac{1}{x^4} = 194 , find x^3 + \frac{1}{x^3} x^2 + \frac{1}{x^2} and x + \frac{1}{x}

Answer:

Given x^4 + \frac{1}{x^4} = 194

\Rightarrow \Big( x^2 + \frac{1}{x^2} \Big)^2 =x^4 + \frac{1}{x^4} + 2

\Rightarrow \Big( x^2 + \frac{1}{x^2} \Big)^2 =196

\Rightarrow \Big( x^2 + \frac{1}{x^2} \Big) =  14

Therefore \Big( x + \frac{1}{x} \Big)^2 =x^2 + \frac{1}{x^2} + 2

\Rightarrow \Big( x + \frac{1}{x} \Big)^2 =16

\Rightarrow \Big( x + \frac{1}{x} \Big) =  \pm 4

If  x + \frac{1}{x} = 4

\Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3. x . \frac{1}{x} (x + \frac{1}{x} )

\Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3 (x + \frac{1}{x} )

x^3 + \frac{1}{x^3} = 4^3 -3(4) = 64-12 = 52

Similarly, if  x + \frac{1}{x} = -4

Then x^3 + \frac{1}{x^3} = (-4)^3 +3(-4) = -52

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(xxx) x - \frac{1}{x} = 3+2\sqrt{2} , find x^3 - \frac{1}{x^3}

Answer:

x - \frac{1}{x} = 3+2\sqrt{2}

\Rightarrow \Big( x - \frac{1}{x} \Big)^3 = x^3 - \frac{1}{x^3} - 3. x . \frac{1}{x} (x - \frac{1}{x} )

\Rightarrow \Big( x - \frac{1}{x} \Big)^3 = x^3 - \frac{1}{x^3} - 3 (x - \frac{1}{x} )

x^3 - \frac{1}{x^3} = (3+2\sqrt{2})^3 +3(3+2\sqrt{2})

= (3+2\sqrt{2}) \Big[ (3+2\sqrt{2})^2 +3 \Big]

= (3+2\sqrt{2})(20+12\sqrt{2}) = 108+76\sqrt{2}

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(xxxi) x = \frac{1}{4-x} , find x + \frac{1}{x} x^3 + \frac{1}{x^3} and x^6 + \frac{1}{x^6}

Answer:

Given x = \frac{1}{4-x}

\Rightarrow x^2+1=4x \Rightarrow x + \frac{1}{x} = 4

\Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3. x . \frac{1}{x} (x + \frac{1}{x} )

\Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3 (x + \frac{1}{x} )

x^3 + \frac{1}{x^3} = 4^3 -3(4) = 64-12 = 52

\Rightarrow \Big( x^3 + \frac{1}{x^3} \Big)^2 = 52

\Rightarrow x^6 + \frac{1}{x^6} + 2 = 2704

\Rightarrow x^6 + \frac{1}{x^6} = 2702

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(xxxii) \Big( x + \frac{1}{x} \Big)^2 = 3 find x^3 + \frac{1}{x^3}

Answer:

\Rightarrow \Big( x + \frac{1}{x} \Big)^2 =3

\Rightarrow \Big( x + \frac{1}{x} \Big) =  \pm \sqrt{3}

If  x + \frac{1}{x} = \sqrt{3}

\Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3. x . \frac{1}{x} (x + \frac{1}{x} )

\Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3 (x + \frac{1}{x} )

x^3 + \frac{1}{x^3} = (\sqrt{3})^3 -3(\sqrt{3}) = 3\sqrt{3} - 3\sqrt{3} = 0

Similarly, if  x + \frac{1}{x} = -\sqrt{3}

Then x^3 + \frac{1}{x^3} = (-\sqrt{3})^3 - 3(-\sqrt{3}) = 0

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(xxxiii) If x + \frac{1}{x} = k prove that x^3 + \frac{1}{x^3} = k(k^2-3)

Answer:

If  x + \frac{1}{x} = k

\Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3. x . \frac{1}{x} (x + \frac{1}{x} )

\Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3 (x + \frac{1}{x} )

x^3 + \frac{1}{x^3} = (k)^3 -3(k) = k(k^2-3) Hence proved.

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(xxxiv) If a+b = 10 and ab=16 , find the value of a^2-ab+b^2 and a^2+ab+b^2

Answer:

Given: a+b = 10 and ab=16

(a+b)^2 = a^2 + 2ab + b^2

\Rightarrow a^2 + ab + b^2 = (a+b)^2 -ab = 10^2 - 16 = 84

also \Rightarrow a^2  - ab + b^2 = (a+b)^2 - 3ab = 10^2 - 3 \times 16 = 100 - 48 = 52

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(xxxv) If a+b = 8 and ab=6 , find the value of a^3+b^3

Answer:

Given: a+b = 8 and ab=6

(a+b)^3 = a^3 + b^3 + 3ab(a+b)

\Rightarrow a^3 + b^3 = (a+b)^3 - 3ab(a+b) = 8^3 - 3 \times 6 \times 8 = 368 

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(xxxvi) If a-b = 6 and ab=20 , find the value of a^3-b^3

Answer:

Given: a-b = 6 and ab=20

(a-b)^3 = a^3 - b^3 - 3ab(a-b)

\Rightarrow a^3 - b^3 = (a-b)^3 + 3ab(a-b) = 6^3 + 3 \times 20 \times 6 = 576 

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Question 5:

(i) If 3x+2y = 12 and xy = 6 , find the value of 9x^2+4y^2

Answer:

(3x+2y)^2 = 9x^2 + 4y^2 + 12xy

\Rightarrow 9x^2 + 4y^2 = 12^2 - 12 \times 6 = 72

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(ii) 4x^2+y^2 = 40 and xy = 6 , find the value of 2x+y

Answer:

(2x+y)^2 = 4x^2 + y^2 + 4xy = 40 + 4 \times 16 = 64

\Rightarrow 2x+y = \pm 8

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(iii) If a^2+b^2+c^2 - ab - bc - ca = 0 , prove that a = b = c

Answer:

Given a^2+b^2+c^2 - ab - bc - ca = 0

\Rightarrow 2a^2+2b^2+2c^2 - 2ab - 2bc - 2ca = 0

\Rightarrow (a^2 -2ab+b^2) +(b^2 -2bc + c^2) + (c^2 -2ca + a^2) = 0

\Rightarrow (a-b)^2+(b-c)^2 + (c-a)^2 = 0

\Rightarrow a-b = 0 \ and \  b -c = 0 \ and \  c-a = 0

\Rightarrow a = b \ and \  b = c \ and \  c= a

\Rightarrow a = b = c

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(iv) If 9x^2+25y^2 = 181 and xy = -6 , find the value of 3x+5y 

Answer:

(3x+5y)^2 = 9x^2 +25y^2 + 30xy = 181-180 = 1 

\Rightarrow 3x+5y = \pm 1 

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(v) If 2x+3y=8 and xy = 2 , find the value of 4x^2+9y^2 

Answer:

(2x+3y)^2 = 4x^2 +9y^2+12xy 

\Rightarrow 4x^2 + 9y^2 = 8^2 - 12 \times 2 = 64 -24 =40 

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(vi) If 3x-7y=10 and xy = -1 , find the value of 9x^2+49y^2 

Answer:

(3x+7y)^2 = 9x^2 + 49y^2 + 42xy 

\Rightarrow 9x^2 + 49y^2 = 10^2 -42(-1) = 142 

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(vii) Prove that a^2+b^2+c^2 - ab - bc - ca = 0 is always non negative for all values of a, b, \ and \ c .

Answer:

To prove that a^2+b^2+c^2 - ab - bc - ca = 0 is always non-negative or To prove that 2a^2+2b^2+2c^2 - 2ab - 2bc - 2ca = 0 is always non-negative

\Rightarrow (a-b)^2 + (b-c)^2 + (c-a)^2 which is always not negative.

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(viii) If a+b = 8 and a-b=6 , find (a^2+b^2) 

Answer:

(a+b)^2 = a^2 + b^2 + 2ab

\Rightarrow a^2 +b^2 = 8^2 - 2 \times 6 = 64-12 = 52

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(ix) If a+b = 6 and a-b=4 , find ab

Answer:

(a+b)^2 = (a-b)^2 +4ab

\Rightarrow 4ab = (a+b)^2 - (a-b)^2 = 36 - 24 = 12

\Rightarrow ab = 3

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(x) If 2x-y+z=0 , prove that 4x^2-y^2+z^2+4xz=0

Answer:

Given: 2x-y+z=0

\Rightarrow 2x+z = y

\Rightarrow 4x^2 + z^2 + 4xz = y^2

\Rightarrow 4x^2-y^2+z^2+4xz=0 $

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(xi) If 2x+3y=13 and xy=6 , find the value of 8x^3+27y^3

Answer:

Given: 2x+3y=13 and xy=6

(2x+3y)^3 = 8x^3+27y^3 + 3 \times 2x \times 3y (2x+3y)

\Rightarrow 8x^3+27y^3 = 13^3 - 3 \times 6 \times 6 \times 13 = 793

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(xii) If 3x-2y=13 and xy = 12 , find the value of 27x^3-8y^3

Answer:

Given: 3x-2y=13 and xy = 12

(3x-2y)^3 = 27x^3-8y^3 - 3 \times 3x \times 2y (3x-2y)

\Rightarrow 27x^3-8y^3 = 13^3 + 3 \times 6 \times 12 \times 13 = 5005

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(xiii) If 4x-5z=16 and xz=12 , find the value of 64x^3-125z^3

Answer:

Given: 4x-5z=16 and xz=12

(4x-5y)^3 = 64x^3-125z^3 - 3 \times 4x \times 5y (4x-5y)

\Rightarrow 64x^3-125z^3 = 16^3 + 3 \times 20 \times 12 \times 16 = 15616

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(xiv) If \frac{x^2+1}{x} = 4 , find the value of 2x^3+ \frac{2}{x^3}

Answer:

Given: \frac{x^2+1}{x} = 4

\Rightarrow x + \frac{1}{x} = 4

x + \frac{1}{x} = 4

\Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3. x . \frac{1}{x} (x + \frac{1}{x} )

\Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3 (x + \frac{1}{x} )

x^3 + \frac{1}{x^3} = 4^3 -3(4) = 64-12 = 52

Therefore 2 \Big( x^3 + \frac{1}{x^3} \Big) = 52 \times 2 = 108

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(xv) If x + y + z = 8 and xy+yz+zx = 20 , find the value of x^3+y^3+z^3-3xyz

Answer:

We know x^3 + y^3 + z^3 -3xyz = (x+y+z)(x^2+y^2+z^2 -xy-yz-zx)

\Rightarrow x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2+2xy+2yz+2zx-3xy-3yz-3zx)

\Rightarrow x^3+y^3+z^3-3xyz=(x+y+z) \Big( (x+y+z)^2-3(xy+yz+zx) \Big)

\Rightarrow x^3 + y^3 + z^3 -3xyz = 8 (8^2 - 3 \times 20) = 8 \times 4 = 32

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