Question 64
If x + y + z = 19, $$x^2 +y^2 + z^2 = 133$$ and $$xz = y^2$$, then the difference between z and x is:
Solution
x + y + z = 19
$$(x + y + z)^2 = x^2 + y^2 +z^2 + 2(xy + yz +zx) $$
361 = 133 + 2(xy + yz +zx)
xy + yz +zx= 114
Substituting xz = y$$^2$$
xy+yz+y$$^2$$ = 114
y(x + y + z) =114
y=6
x+z=13
xz=$$y^2$$ = 36
x - z = $$\sqrt{ (x + z)^2 - 4xz}$$
= $$\sqrt{ 169 - 144}$$
=5
So , the answer would be Option a)5.
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