Algebra
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Q. if $$x+y+z=19$$, $$x^2+y^2+z^2=133$$ and $$xz=y^2$$, then the difference between $$z$$ and $$x$$ is:

  • (A). 3
  • (B). 4
  • (C). 6
  • (D). 5
Option (D) is Correct

$$(x+y+z)^2$$ = $$x^2+y^2+z^2+2(xy+yz+zx)$$
$$19^2=133+2(xy+yz+y^2)$$
$$2y(x+y+z)=361-133$$
$$2yxx19=228$$
$$y=228/38=6$$
Now,
$$x+z=19-y$$
$$x+z=19-6=13$$
$$x-z=sqrt((x+z)^2-4xz)$$
$$=sqrt(13^2-4y^2)$$
$$=sqrt(169-144)=5$$