Question

$if \: x = \frac{1}{2 - \sqrt{3} } \: show \: value \: of \: x {}^{3} - 2 {x}^{2} - 7x + 5 \: is \: 3$
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Answer 1

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Given ✍ :

$\bold{If\:x\:=\: \frac {1}{2\:-\: \sqrt {3}}}$

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➾Required to find :

1. $\bold{Value\:of\:{x}^{3}\:-\:2{x}^{2}\:-\:7x\:+\:5\:=\:3}$

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◈Rationalising the denominator :

Rationalising the denominator is important before Solving any question .

So; similarly we have to Rationalise the denominator of the value given for " x "

The value of x is

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

However;

we have to Rationalise the denominator of value of x

So;

Rationalising Factor of 2 - √3 is 2 + √3

Here is the process of Rationalising.

$\frac{1}{2 - \sqrt{3} } \\ \\ = \frac{1}{2 - \sqrt{3} } \times \frac{2 + \sqrt{3} }{2 + \sqrt{3} } \\ \\ = \frac{2 + \sqrt{3} }{(2 {)}^{2} - ( \sqrt{3} {)}^{2} } \\ \\ = \frac{2 + \sqrt{3} }{4 - 3} \\ \\ = \frac{2 + \sqrt{3} }{1}$

$\boxed{\longrightarrow{\large{\boxed {value\:of\:x\:is\:2\:+\: \sqrt {3}}}}}$

Successful , the value of x is 2 + √3 .

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Solution ✏ :

From the above we came to know that;

Value of x is 2 + √3 .

Now we have to find the value of x^3 and x^2 because they have powers on them .

So ;

Value of x^3 is ;

$(2 + \sqrt{3} {)}^{3} \\ using \: (x + y {)}^{3} = {x}^{3} + {y}^{3} + 3 {x}^{2} y + 3x {y}^{2} \\ so \\ (2 + \sqrt{3} {)}^{3} \\ \\ = (2 {)}^{3} + ( \sqrt{3} {)}^{3} + 3(2 {)}^{2} ( \sqrt{3} ) + 3(2)( \sqrt{3} {)}^{2} \\ \\ = 8 + 3 \sqrt{3} + 12 \sqrt{3} + 18 \\ \\ = 26 + 15 \sqrt{3}$

Similarly;

We have to find the Value of x^2

That is ;

${x}^{2} \\ = (2 + \sqrt{3} {)}^{2} \\ using \: (x + y {)}^{2} = {x}^{2} + {y}^{2} + 2xy \\ \\ = (2 {)}^{2} + ( \sqrt{3} {)}^{2} +2 (2)( \sqrt{3} ) \\ \\ = 4 + 3 + 4 \sqrt{3} \\ \\ = 7 + 4 \sqrt{3}$

Hence ;

Now let's find the value of full expressions

That is

${x}^{3} - 2 {x}^{2} - 7x + 5$

Hence; substitute the values.

$26 + 15 \sqrt{3} - 2(7 + 4 \sqrt{3} ) - 7(2 + \sqrt{3} ) + 5 \\ \\ 26 + 15 \sqrt{3} - 14 - 8 \sqrt{3} - 14 - 7 \sqrt{3} + 5 \\ \\ 26 - 14 - 14 + 5 + 15\sqrt{3} - 8 \sqrt{3} - 7 \sqrt{3} \\ \\ 31 - 28 + 15 \sqrt{3} - 15 \sqrt{3} \\ \\ 3 + 0 \\ \\ = 3$

Therefore;

The value of

$value \: of \: {x}^{3} - 2 {x}^{2} - 7x + 5 \: is \: 3$

✔ Hence proved

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Answer 2

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✰Answer✰

♦️GiveN:

• $\large{ \tt{x = \frac{1}{2 - \sqrt{3} } }}$

♦️To ProvE:

• $\large{ \tt{ \: {x}^{3} - 2 {x}^{2} - 7x + 5 = \: \: 3}}$

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✰Solution✰

➯First we need to rationalize the value of x , So that it could be easier to use in further calculation.

☛Rationalizing the given value of x,

$\large{ \tt{ \rightarrow \: x = \frac{1}{2 - \sqrt{3} } }} \\ \\ \large{ \sf{ \star{ \: rationalizing \: factor \: of \: 2 - \sqrt{3} = 2 + \sqrt{3} }}} \\ \\ \large{ \tt{ \rightarrow \: x = \frac{1(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3} ) } }} \\ \\ \large{ \tt{ \rightarrow \: x = \frac{2 + \sqrt{3} }{ {2}^{2} - { (\sqrt{3}) }^{2} } }} \\ \sf{ \green{by \: using \: (a + b)(a - b) = {a}^{2} - {b}^{2} }} \\ \\ \large{ \tt{ \rightarrow \: x = \frac{2 + \sqrt{3} }{4 - 3} }} \\ \\ \large{ \tt{ \rightarrow \: x = \frac{2 + \sqrt{3} }{1} }} \\ \\ \large{ \rightarrow{ \boxed{ \purple{ \tt{x = 2 + \sqrt{3} }}}}}$

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➯We had our equation to prove:

$\large{ \tt{ \star{ \: \: \: {x}^{3} - 2 {x}^{2} - 7x + 5 \: = \: 3}}}$

⭐Without putting actual value of x , Let's break the to proof equation down in such a way that at last we get our value by simplification only.....

♦️First of all, forming an equation with x

➯We have,

$\large{ \tt{ \rightarrow \: x = 2 + \sqrt{3} }}$

➯Subtracting 2 from both sides,

$\large{ \tt{ \rightarrow \: x - 2 = \sqrt{3} }}$

➯Squaring both sides,

$\large{ \tt{ \rightarrow \: (x - 2) {}^{2} = 3}} \\ \\ \large{ \tt{ \rightarrow \: {x}^{2} - 4x + 4 = 3}}$

➯Subtracting 3 both sides,

$\large{ \rightarrow \: \boxed{ \tt{{x}^{2} - 4x + 1 = 0}}}.......(1)$

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✰Proof✰

$\large{ \sf{ \underline{ \underline{LHS}}}}$

$\large{ \tt{ \rightarrow \: {x}^{3} - 2 {x}^{2} - 7x + 5}}$

➯Arranging to get equation (1)

$\large{ \tt{ \rightarrow \: x( {x}^{2} - 4x + 1) + 2 {x}^{2} - 8x + 5}}$

➯Putting value of equation (1),

$\large{ \tt{ \rightarrow \: x(0) + 2 {x}^{2} - 8x + 5}} \\ \\ \large{ \tt{ \rightarrow \: 2 {x}^{2} - 8x + 5}}$

➯Again arranging for equation (1),

$\large{ \tt{ \rightarrow \: 2( {x}^{2} - 4x + 1) + 3}}$

➯Putting value of equation (1),

$\large{ \tt{ \rightarrow \: 2(0) + 3}} \\ \\ \large{ \tt{ \rightarrow \: 3}} \: \: \: \: \: \: \large{ \sf{ \underline{ \underline{(RHS)}}}}$

⭐ Hence,proved!!!!

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