Question

Solve :
$x^{2002} + 10x^{2001} \div 10x^{2000} = 957.9$
​

Answer 1

Answer:

x = -103 x = 93

Step-by-step explanation:

We have,

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

To find, the value of x = ?

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

⇒ [tex]x^{2002} \div 10x^{2000} + [tex]10x^{2000} \div 10x^{2000}

⇒{ x^2002-200}/10+{x^2001-2000}=957.9

⇒ x²/10+x=957.9

⇒( x²+10x) /10=957.9

By crossmultiplication, we get

⇒ x²+10x=957.9×10

⇒ x²+10x=9579

⇒ x²+ 10x - 9579 = 0

⇒x² + 103x - 93x - 9579 = 0

⇒ x(x + 103) - 93(x + 103) = 0

⇒ (x - 93)(x + 103) = 0

⇒ x - 93 = 0 or, x + 103 = 0

⇒ x = 93  or, x = - 103

∴ x = 93  or, x = - 103