find the value of x in this question
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Sachinhero11:
kya hua??
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Heya User,
--> Sorry for the delay but it took me quite long to write a satisfactory solution to this ..
--> However, since I have shortage of time ^_^ , I'll provide the basic steps for now, and will upload my answer to the satisfactory level later..
--> [ x² + [ ax / ( a + x ) ]² ] = 3a²
--> Expanded as :->
==> - 3a⁴ - 6a³x - a²x² + 2ax³ + x⁴ = 0
== 3a⁴ + 6a³x + a²x² - 2ax³ - x⁴ = 0
== [ 3a⁴ + 3a³x + 3a³x - 3a²x² + 3a²x² + a²x² - 3ax³ + ax³ - x⁴ ] = 0
== [ 3a⁴ + 3a³x - 3a²x² + 3a³x + 3a²x² - 3ax³ + a²x² + ax³ - x⁴ ] = 0
== [ 3a⁴ + 3a³x - 3a²x² ] + [ 3a³x + 3a²x² - 3ax³ ] + [ a²x² + ax³ - x⁴ ] = 0
== 3a² [ a² + ax - x² ] + 3ax [ a² + ax - x² ] + x² [ a² + ax - x² ] = 0
== [ a² + ax - x² ] [ 3a² + 3ax + x² ] = 0
However, the second factor gives an imaginary soln..
--> Solve for 'x' in the first factor, to get :->
-->
And hence, you have the answer in option ( ii ) and ( iv ) ^_^
+_+ I'll provide uh with a comparatively easy and appropriate soln. pretty soon ... ATB
^_^
--> Sorry for the delay but it took me quite long to write a satisfactory solution to this ..
--> However, since I have shortage of time ^_^ , I'll provide the basic steps for now, and will upload my answer to the satisfactory level later..
--> [ x² + [ ax / ( a + x ) ]² ] = 3a²
--> Expanded as :->
==> - 3a⁴ - 6a³x - a²x² + 2ax³ + x⁴ = 0
== 3a⁴ + 6a³x + a²x² - 2ax³ - x⁴ = 0
== [ 3a⁴ + 3a³x + 3a³x - 3a²x² + 3a²x² + a²x² - 3ax³ + ax³ - x⁴ ] = 0
== [ 3a⁴ + 3a³x - 3a²x² + 3a³x + 3a²x² - 3ax³ + a²x² + ax³ - x⁴ ] = 0
== [ 3a⁴ + 3a³x - 3a²x² ] + [ 3a³x + 3a²x² - 3ax³ ] + [ a²x² + ax³ - x⁴ ] = 0
== 3a² [ a² + ax - x² ] + 3ax [ a² + ax - x² ] + x² [ a² + ax - x² ] = 0
== [ a² + ax - x² ] [ 3a² + 3ax + x² ] = 0
However, the second factor gives an imaginary soln..
--> Solve for 'x' in the first factor, to get :->
-->
And hence, you have the answer in option ( ii ) and ( iv ) ^_^
+_+ I'll provide uh with a comparatively easy and appropriate soln. pretty soon ... ATB
^_^
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