x^2 + 6y=7 and y=x^3 . find the value of x,y.
Answers
Answered by
1
x^2 + 6x^3 = 7
x^2 (6x + 1 ) =7
x^2 = 7 and 6x +1 =7
X=√7 and X = 5/6
y = √7^3 and y = (5/6) ^3
y= 49√7 and y= 125/216 ......... ya it's correct dear
x^2 (6x + 1 ) =7
x^2 = 7 and 6x +1 =7
X=√7 and X = 5/6
y = √7^3 and y = (5/6) ^3
y= 49√7 and y= 125/216 ......... ya it's correct dear
jiaurrahamankhanjrk:
no brother, the answet is x=y=1. so i think u r wrong
bt watch my soln. their is no mistake yarr...
so how ??
r u on Facebook? here its difficult to expalin mistake. i want to send pic
no...i dont use fb ...
no prob... I'll see it
ok,
bt how, here it is impossible to send pic
Answered by
3
Heya User,
--> Umm, I'm lazy at sending pics =_=
--> x² + 6y = 7 || y = x^3
We turn it to an eqn. in x ;
--> x² + 6 x^3 = 7
--> 6x³ + x² - 7 = 0
--> ( x - 1 )( 6x² + 7x + 7 ) = 0
--> ( x - 1 ) = 0 ; or --> ( 6x² + 7x + 7 ) = 0
--> x = 1 or --> x = [ - 7 ± √119 i ]/12
=_= However, I'm willing toh avoid the latter value for 'x' coz tht's 'C'
=> x = 1 || y = x³ = 1 is the soln. :)
--> Umm, I'm lazy at sending pics =_=
--> x² + 6y = 7 || y = x^3
We turn it to an eqn. in x ;
--> x² + 6 x^3 = 7
--> 6x³ + x² - 7 = 0
--> ( x - 1 )( 6x² + 7x + 7 ) = 0
--> ( x - 1 ) = 0 ; or --> ( 6x² + 7x + 7 ) = 0
--> x = 1 or --> x = [ - 7 ± √119 i ]/12
=_= However, I'm willing toh avoid the latter value for 'x' coz tht's 'C'
=> x = 1 || y = x³ = 1 is the soln. :)
thanks bro, its is absolutely right. THANKS A LOT
Eh! =_=
Hit the Thanks button =_=
xD :)
already hit
Similar questions