Question

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

Answer 1

Answer:

$\sqrt{x^2+y^2+1}=18$

Step-by-step 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:

$\sqrt{x^2+y^2+1}=\sqrt{324}=18$

Thus:

$\mathbf{\sqrt{x^2+y^2+1}=18}$