Question

Express as the sum of perfect square . x² + y² + z² + xy + yz - zx . Question - (2) if x - 1/x = 3/2. then find x² - 1/x² = ? x⁴ + 1/x⁴ = ? need verified Answer not spammed.​

Answer 1

Question - ( i ) :-

Given expression :-

• x² + y² + z² + xy + yz - zx

Solution and concept :-

To solve this expression at first we have to factories by applying identity.

Calculation begins :-

⟶ x² + y² + z² + xy + yz - zx

• Here we have to add something but expression will be same without changing its expression.

↠ 1/2 *[2(x² + y² + z² + xy + yz - zx)]

↠ 1/2[2x² + 2y² + 2z² + 2xy + 2yz - 2zx]

↠1/2[x² + x² + y² + y² + z² + z² + 2xy + 2yz - 2zx]

• (a + b)² = a² + b² + 2ab
• (a - b)² = a² + b² - 2ab

↠1/2[(x² + y² + 2xy) + (y² + z² + 2yz) + (z² + x² - 2zx)]

↠1/2[(x + y)² + (y + z)² + (z - x)²]

Hence,

↠The required answer for this expression:-

• ↠1/2[(x + y)² + (y + z)² + (z - x)²]

Question - ( ii ) :-

Given :-

• x - 1/x = 3/2

To Find :

• x² - 1/x² = ? x⁴ + 1/x⁴ = ?

Solution :-

• To calculate value of above expression at first we have to find the value of x by simplifying the given expression.

Calculation begins :-

↠x - 1/x = 3/2

↠x² - 1/x = 3/2

↠2x² - 2 = 3x

↠2x² - 3x - 2 = 0

• Splitting the middle term :-

↠2x² - 4x + x - 2 = 0

↠2x(x - 2) + 1(x - 2) = 0

↠(2x + 1)(x - 2) = 0

↠ Therefore, x = -1/2 or 2

• Here we take x = 2 because negative value is not acceptable for the given expression.

Value of x² - 1/x² =?

• By putting x = 2 we get :-

↠2² - 1/2² ↠4 - 1/4 ↠16 - 1/4 ↠ 15/4

↠x² - 1/x² = 15/4

Value of x⁴ + 1/x⁴ = ?

• By putting x = 2 we get :-

↠2⁴ + 1/2⁴ ↠ 16 + 1/16 ↠ 256 + 1/16 ↠ 257/16

↠x⁴ + 1/x⁴ = 257/16

Answer 2

First answer:

Information given to us:

• x² + y² + z² + xy + yz - zx

We have to calculate:

• We have to express them as the sum of perfect square

Calculations we are performing:

$\red \bigstar \: \underline{ \sf{multiplying \: \dfrac{1}{2} \: with \: x {}^{2} + y {}^{2} + z {}^{2}:- \ }}$

$: \large{ \longmapsto} \: \: \dfrac{1}{2} \: ( \sf{ \:x {}^{2} \: + \: y {}^{2} \: + xy \: + yz \: - \: zx) }$

$\red \bigstar \: \underline{ \sf{ Also \: multiplying \: 2 \: with \: x {}^{2} + y {}^{2} + z {}^{2}:- \ }}$

$: \large{ \longmapsto} \: \: \: \sf{ 2 \:x {}^{2} \: + \: 2y {}^{2} \: + 2xy \: + 2yz \: - \: 2zx }$

$\red \bigstar \: \underline{ \sf{Splitting \: terms:- \ }}$

$: \large{ \longmapsto} \: \: \: \sf{ \: x {}^{2} \: + x {}^{2} \: + y {}^{2} \: + y {}^{2} \: + z {}^{2} \: + z {}^{2} + \: 2xy \: + \: 2zx}$

$: \large{ \longmapsto} \: \: \: \sf{ 2 \:x {}^{2} \: + \: 2y {}^{2} \: + 2xy \: + 2yz \: - \: 2zx }$

$\red \bigstar \: \underline{ \sf{By \: using \: this \: identity \: we would \: get:- \ }}$

$: \implies \sf{(a + b) {}^{2} \: = \: a {}^{2} \: + \: 2ab \: + b {}^{2} }$

$\large{ \longmapsto} \: \sf{ \dfrac{1}{2} (\: x + y {}^{2} + \: 2xy) }$

$\large{ \longmapsto} \: \sf{ \dfrac{1}{2} \: [(\: x + y {}^{2} + \: 2xy \: )(y {}^{2} \: + \: z {}^{2} \: + 2yz) ] }$

And,

$\red \bigstar \: \underline{ \sf{By \: using \: this \: identity \: we would \: get:- \ }}$

$: \implies \sf{(a - b) {}^{2} \: = \: a {}^{2} \: - 2ab \: + b {}^{2} }$

$\large{ \longmapsto} \: \sf{ \dfrac{1}{2} \: (z {}^{2} \: + \: x {}^{2} \: - \: 2zx)}$

$\red \bigstar \: \underline{ \sf{Now \: adding \: all \: of \: them :- \ }}$

$\large{ \longmapsto} \: \sf{ \dfrac{1}{2} \: (x {}^{2} \: + \: y {}^{2} \: + \: 2xy) \: + \: (y {}^{2} \: + \: z {}^{2} \: + \: 2yz \: + \: (z {}^{2} \: + \: x {}^{2} \: - \: 2zx) }$

$\red \bigstar \: \underline{ \sf{Therefore, \: sum \: of \: perfect \: squares \: would \: be :- \ }}$

$: \large{ \longmapsto} \: \sf{ \dfrac{1}{2} \: [ \: (x \: + y \: ) {}^{2} + \: (y \: + z) {}^{2} \: + \: (z \: - \: x) {}^{2} ] }$

_________________

Second answer:

Information given to us:

• $\large{ \sf{x \: - \: \dfrac{x}{1} \: = \: \dfrac{3}{2} }}$

What we have to calculate:-

• $\large{ \sf{x {}^{2} \: - \: \dfrac{1}{x {}^{2} } \: \: \dfrac{{} }{} }}$
• $\large{ \sf{x {}^{4} \: + \: \dfrac{1}{x {}^{4} } \: \: \dfrac{{} }{} }}$

Performing calculations:-

$\red \bigstar \: \underline{ \sf{First \: we \: are \:finding \: out \: value \: of \: the \: variable \: x:- }}$

$: \large{ \longmapsto} \: \sf{x \: - \: \dfrac{1}{x} \: = \: \dfrac{3}{2} }$

$\red \bigstar \: \underline{ \sf{Taking \: L.CM.:- }}$

$: \large{ \longmapsto} \: \sf{x {}^{2} \: - \: \dfrac{1}{x} \: = \: \dfrac{3}{2} }$

$\red \bigstar \: \underline{ \sf{Acq,:- }}$

$: \large{ \longmapsto} \: \sf{2x {}^{2} \: - \: 2 \: = \: 3x }$

$\red \bigstar \: \underline{ \sf{Changing \: the \: sides \: positive \: term \: would \: be \: negative:- }}$

$: \large{ \longmapsto} \: \sf{2x {}^{2} \: - \: 3x \: - 2 \: = \: 0 }$

Now,

$: \large{ \leadsto} \: \sf{2x {}^{2} - 4x + x - 2 \: = \: 0 } \\ \large{ : \: \leadsto} \: \sf{2x(x - 2) \: + 1(x - 2)} \\ \large{ : \leadsto} \sf{} \: (2x + 1) \: (x - 2)$

$\red \bigstar \: \underline{ \sf{Hence, \: value \: x \: would \: be \: 2 :- }}$

As negative integers can't be accepted.

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