Question

If x ≠ y and,
→ x² = 17x + y
→ y² = x + 17y

Find the value of √(x² + y² + 1)​

Answer 1

Answer Keys.

We can add or subtract both sides of an equation by an equal value. If we subtract the equation we will find a solution.

Solution.

Let's subtract the equation.

$x^2-y^2=16x-16y$

$\rightarrow (x+y)(x-y)=16(x-y)$

$\rightarrow (x+y-16)(x-y)=0$

$\therefore x+y=16$

Then, let's add the equation.

$x^2+y^2=18x+18y$

$\rightarrow x^2+y^2=18\cdot 16$

$\rightarrow x^2+y^2=17^2-1^2$

So, our answer is

$\sqrt{x^2+y^2+1} =\sqrt{17^2} =\boxed{17}$

More information.

What is the value of $xy$ here?

Let $\begin{cases} & x+y=u\\ & xy=v\end{cases}$

Then $x^2+y^2=u^2-2v$

$\rightarrow 17^2-1^2=16^2-2v$

$\rightarrow 2v=16^2-17^2+1^2$

$\rightarrow 2v=-33+1$

$\rightarrow 2v=-32$

$\therefore v=-16$

So, the value of $xy$ is $-16$.

Answer 2

Adding the two given equations, we get,

${x}^{2} + {y}^{2} =18(x+y).$

Subtracting the second equation from the first, we get,

${x}^{2} − {y}^{2} =16(x−y).$

⇒(x+y) (x−y) = 16 (x−y).

⇒x + y=16.

$⇒ {x}^{2} + {y}^{2} =18×16.$

$⇒ \sqrt{ {x}^{2} + {y}^{2} +1}= \sqrt{18×16+1}$

$= \sqrt{289} = 17.$