2002
2001
solve: x
+ 10%
200
102
952
Answers
Step-by-step explanation:
We have,
\dfrac{x^{2002}+10x^{2001}}{10x^{2000}} =957.9
10x
2000
x
2002
+10x
2001
=957.9
To find, the value of x = ?
∴ \dfrac{x^{2002}+10x^{2001}}{10x^{2000}} =957.9
10x
2000
x
2002
+10x
2001
=957.9
⇒ \dfrac{x^{2002}}{10x^{2000}}+\dfrac{10x^{2001}}{10x^{2000}} =957.9
10x
2000
x
2002
+
10x
2000
10x
2001
=957.9
⇒ \dfrac{x^{2002-2000}}{10}+x^{2001-2000}=957.9
10
x
2002−2000
+x
2001−2000
=957.9
⇒ \dfrac{x^{2}}{10}+x=957.9
10
x
2
+x=957.9
⇒ \dfrac{x^{2}+10x}{10}=957.9
10
x
2
+10x
=957.9
By crossmultiplication, we get
⇒ x^{2}+10x=957.9\times 10=9579x
2
+10x=957.9×10=9579
⇒ x^{2}x
2
+ 10x - 9579 = 0
⇒ x^{2}x
2
+ 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