(x)2-2kx+ (7k-12)=0the value of k such that the equation has real and equal roots k
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Answer:
u
x
=
dx
du
=
x
3
+y
3
+z
3
−3xyz
3x
2
−3yz
..........(1)
u
y
=
dy
du
=
x
3
+y
3
+z
3
−3xyz
3y
2
−3xz
..........(2)
u
z
=
dz
du
=
x
3
+y
3
+z
3
−3xyz
3z
2
−3xy
..........(3)
Add all the three equations
dx
du
+
dy
du
+
dz
du
=
x
3
+y
3
+z
3
−3xyz
3(x
2
+y
2
+z
2
−xy−yz−zx)
=
x+y+z
3
∴(x+y+z)(u
x
+u
y
+u
z
)=3
Note: z
3
+b
3
+c
3
−3abc=(a+b+c)(a
2
+b
2
+c
2
−ab−bc−ca)
⇒x
3
+y
3
+z
3
−3xyz=(x+y+z)(x
2
+y
2
+z
2
−xy−xz−yz)
so, (x+y+z)(u
x
+u
y
+u
z
)=3
Step-by-step explanation:
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