Question

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Question :

Solve :
$\sf{\dfrac{x^{2002}\:+\:10x^{2001}}{10x^{2000}}}$ = 957.9

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Answer 1

Answer:

x = 93,-103

Step-by-step explanation:

$\frac{x^{2002} + 10x^{2001}}{10x^{2000}} = 957.9$

$\Rightarrow \frac{x^{2001}(x + 10)}{10x^{2000}} = 957.9$

$\Rightarrow \frac{x^{2001 - 2000}(x + 10)}{10} = 957.9$

$\Rightarrow \frac{x(x + 10)}{10} = 957.9$

$\Rightarrow x(x + 10) = 9579$

$\Rightarrow x^2 + 10x - 9579 = 0$

$\Rightarrow x^2 + 103x - 93x - 9579 = 0$

$\Rightarrow x(x + 103) - 93(x + 103) = 0$

$\Rightarrow \boxed{x = 93, x = -103}$

Hope it helps!

Answer 2

$\huge\underline\mathfrak\red{Answer-}$

$\huge\boxed{x\:=\:93\:or\:-103}$

$\huge\underline\mathfrak\red{Explanation-}$

$\sf{\dfrac{x^{2002}\:+\:10x^{2001}}{10x^{2000}}}$ = 957.9

Multiply both sides by 10,

$\leadsto$ $\sf{\dfrac{x^{2002}\:+\:10x^{2001}}{10x^{2000}}}$ × 10 = 957.9 × 10

$\leadsto$ $\sf{\dfrac{x^{2002}\:+\:10x^{2001}}{x^{2000}}}$ = 9579

$\leadsto$ $\sf{\dfrac{x^{2002}}{x^{2000}}}$ + $\sf{\dfrac{10x^{2001}}{x^{2000}}}$ = 9579

Now, by using law of exponents :

★ $\sf{\dfrac{a^{m}}{a^{n}}}$ = $\sf{a^{m-n}}$

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$\leadsto$ $\sf{x^{2002-2000}}$ + $\sf{10x^{2001-2000}}$ = 9579

$\leadsto$ $\sf{x^{2}\:+\:10x\:=\:9579}$

$\leadsto$ $\sf{x^{2}\:+\:10x\:-\:9579\:=\:0}$

Now, by splitting middle term,

$\leadsto$ x² + 103x - 93x - 9579 = 0

$\leadsto$ x( x + 103 ) - 93( x + 103 ) = 0

$\leadsto$ ( x - 93 )( x + 103 ) = 0

$\leadsto$ $\huge\boxed{x\:=\:93\:or\:-103}$

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