In Square Root 11-x then plus.... In square root 6-x then = in square root 27-x ............ Using quadratic equation, find answer

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user156724's profile pic

Posted on

`sqrt(11-x)+sqrt(6-x)=sqrt(27-x)`                    (i)

squaring both side.

`11-x+6-x+2sqrt((11-x)(6-x))=27-x`

`17-27-2x+x=-2sqrt((11-x)(6-x))`

`-10-x=-2sqrt((11-x)(6-x))`

 

Squaring agin both side ,we have

`100+x^2+20x=4(11-x)(6-x)`

`100+x^2+20x=4(66-17x+x^2)`

`4x^2-x^2-68x-20x+264-100=0`

`3x^2-88x+164=0`

a=3,b=-88,c=164

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`x=(88+-sqrt(7744-12xx164))/(2xx3)`

`x=(88+-sqrt(5776))/6`

`x=(88+-76)/6`

`x=82/3,2`

`x=82/3 `  will not satisfy equation (i),so it is not possible.

Thus x=2 is  Ans.

Sorry it is minus not plus.. please solve again

user156724's profile pic

Posted on

In Square Root 11-x then minus.... In square root 6-x then = in square root 27-x ............ Using quadratic equation, find answer



Sorry it is minus not plus.. please solve again

user156724's profile pic

Posted on

                    (i)

squaring both side.

 

Squaring agin both side ,we have

a=3,b=-88,c=164

  will not satisfy equation (i),so it is not possible.

Thus x=2 is  Ans.

 

Thank you so much..

pramodpandey's profile pic

Posted on

`sqrt(11-x)+sqrt(6-x)=sqrt(27-x)`                    (i)

squaring both side.

`11-x+6-x+2sqrt((11-x)(6-x))=27-x`

`17-27-2x+x=-2sqrt((11-x)(6-x))`

`-10-x=-2sqrt((11-x)(6-x))`

 

Squaring agin both side ,we have

`100+x^2+20x=4(11-x)(6-x)`

`100+x^2+20x=4(66-17x+x^2)`

`4x^2-x^2-68x-20x+264-100=0`

`3x^2-88x+164=0`

a=3,b=-88,c=164

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`x=(88+-sqrt(7744-12xx164))/(2xx3)`

`x=(88+-sqrt(5776))/6`

`x=(88+-76)/6`

`x=82/3,2`

`x=82/3 `  will not satisfy equation (i),so it is not possible.

Thus x=2 is  Ans.

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