1096175
contestada

The real numbers $x$ and $y$ are such that \begin{align*} x + y &= 4, \\ x^2 + y^2 &= 22, \\ x^4 &= y^4 - 176 \sqrt{7}. \end{align*}Compute $x - y.$

Respuesta :

LammettHash

You get everything you need from factoring the last expression:

[tex]x^4-y^4=-176\sqrt7[/tex]

The left side is a difference of squares, and we get another difference of squares upon factoring. We end up with

[tex]x^4-y^4=(x^2-y^2)(x^2+y^2)=(x-y)(x+y)(x^2+y^2)[/tex]

Plug in everything you know and solve for [tex]x-y[/tex]:

[tex]-176\sqrt7=(x-y)\cdot4\cdot22\implies x-y=\boxed{-2\sqrt7}[/tex]

Answer:

-2sqrt(7)

Step-by-step explanation:

Solution:

From the third equation, $x^4 - y^4 = -176 \sqrt{7}.$

By difference of squares, we can write

\[x^4 - y^4 = (x^2 + y^2)(x^2 - y^2) = (x^2 + y^2)(x + y)(x - y).\]Then $-176 \sqrt{7} = (22)(4)(x - y),$ so $x - y = \boxed{-2 \sqrt{7}}.$

Otras preguntas