If x+y+z=5, x
2
+y
2
+z
2
=21 and y
2
=zx, then the value of y is:

Given equations:
x+y+z=5⟹x+z=5−y

x
2
+y
2
+z
2
=21⟹x
2
+z
2
=21−y
2


y
2
=zx

We know that:

(x+z)
2
=x
2
+z
2
+2zx
Substitute the values from the modified equations into this identity:

(5−y)
2
=(21−y
2
)+2(y
2
)
25−10y+y
2
=21−y
2
+2y
2

25−10y+y
2
=21+y
2

Subtract y
2
from both sides:

25−10y=21
10y=25−21
10y=4
y=
10
4
​
=
5
2
​