Question
If (x2Â + y2Â + z2Â -
4x + 6y + 13) = 0, then find the value of (x + y + z).Solution
We can re-write the above expression as:
x2Â - 4x + 4 + y2Â + 6y + 9 + z2Â = 0
Or, (x - 2)2Â + (y + 3)2Â + z2Â = 0
Since, square of a number can never be negative, all (x - 2)2, (y + 3)2Â and z2Â have to be '0'.
So, x - 2 = 0
Or, x = 2
(y + 3)2Â = 0
And, y = -3
And, z2Â = 0
Or, z = 0
So, required value = 2 - 3 + 0 = -1
If a+1/a =9, then a5+1/a5 is equal to:
If a + (1/a) = 2, then find the value of (a5 + a3 + 6)/(7a – 5).Â
22% of ‘x’ is equal to 66% of ‘y’, then find the value of x: y.
If 3y-3/y=18 then find the value of y² +1/y²
If AB = x + 3, BC = 2x and AC = 4x − 5, then for what value of ‘x’ does B lie on AC?
Solve for x:
(2x − 1)/3 − (x − 4)/2 = 3
If √b + (1/√b) = 11, then find the value of b + (1/b).
If, 4x2 + y2 + 12x + 6y + 18 = 0, then find the value of (2x + y)/(y + 6x).

If ( p = 40 - q - r ) and ( pq + r(q + p) = 720 ), then find the value of ( p2 + q2 + r2).