Question

78. If 5(x2 + y2 + z) = 4(xy + yz + zx) then
X:Y:Z=
(1) 8:1:4 (2) 2:1:8
(3) 4:1:8 (4) 4:8:1​

Answer 1

Given equations
xy+yz+zx=7
⇒xy+(x+y)z=7 .....(1)

x+y+z=6
⇒z=6−(x+y) .....(2)

Substituting the value from eq(2) in eq(1), we get
xy+(x+y){6−(x+y)}=7
⇒xy+6x+6y−x
2
−y
2
−2xy=7
⇒x
2
+y
2
+xy−6x−6y+7=0
⇒x
2
+x(y−6)+(y
2
−6y+7)=0

Since, x is real
⇒D≥0
⇒(y−6)
2
−4(y
2
−6y+7)≥0
⇒y
2
+36−12y−4y
2
+24y−28≥0
⇒−3y
2
+12y+8≥0
⇒3y
2
−12y−8≤0 ....(3)

Now, taking 3y
2
−12y−8=0, we will find the roots
y=
6
12±
144+96
​

​


y=
6
12±
240
​

​


y=
6
12±4
15
​

​


y=
3
6±2
15
​

​


So, the inequality (3) becomes
[y−(
3
6−2
15
​

​
)][y−(
3
6+2
15
​

​
)]≤0

⇒
3
6−2
15
​

​
≤y≤
3
6+2
15
​

​


Since, x,y,z are symmetrical placed . So, y,z also lies in thesame interval.