Question

If x2=18x+y and y2=18y+x, then find the value of root

Answer 1

`√(x^2+y^2+1)=18`

Explanation:

We are given:

`x^2=18x+y\qquad\qquad[1]`

`y^2=18y+x\qquad\qquad[2]`

Subtracting [1] and [2]:

`x^2-y^2=18x+y-(18y+x)`

Operating:

`x^2-y^2=18x+y-18y-x`

Recall:

`x^2-y^2=(x-y)(x+y)`

Substituting:

`(x-y)(x+y)=18x+y-18y-x`

Rearranging:

`(x-y)(x+y)=18x-18y-(x-y)`

`(x-y)(x+y)=18(x-y)-(x-y)`

Dividing by x-y (recall x≠y):

`x+y=18-1=17`

`x+y=17\qquad[3]`

Now we add [1] and [2]:

`x^2+y^2=18x+y+18y+x`

Rearranging:

`x^2+y^2=18x+18y+x+y`

`x^2+y^2=18(x+y)+(x+y)`

`x^2+y^2=19(x+y)`

Substituting from [3]

`x^2+y^2=19*17=323`

Adding 1:

`x^2+y^2+1=324`

Taking the square root:

`√(x^2+y^2+1)=√(324)=18`

Thus:

`\mathbf{√(x^2+y^2+1)=18}`