Question
I: x2Â + 31x + 228 = 0 II:
y2 + 3y – 108 = 0 Direction: In each of the following question, two equations are given. You have to solve both the equations to find the relation between x and y.Solution
x2 + 31x + 228 = 0 x2 + 19x + 12x + 228 = 0 x(x + 19) + 12(x + 19) = 0 (x + 12)(x + 19) = 0 x = -12, -19 y2 + 3y – 108 = 0 y2 + 12y – 9y – 108 = 0 y(y + 12) – 9(y + 12) = 0 (y – 9)(y + 12) = 0 y = 9, -12 x ≤ y
If x + y + z = 20, x² + Y² + z² = 160 and x z = y², then find the value of x z?
if a2 + b2 + c2Â =Â 2(3a -5b -6c)-70 , then a-b-c = ?
If sin 28° = 15/17 , then tan 62° = ?
If ab = -15 and (a2 + b2) = 34, then find the value of (a – b).
(123×123×123 + 130×130×130)/(123×123 - 123×130 + 130×130) = ?Â
 = ?
(√ (sec2 θ + cosec2 θ )) ((sinθ (1 + cosθ ))/(1 + cos&thet...
If z + (1/z) = 2 then find z105 + 1/(z105) = ?
- If (x + y + z) = 14 and (xy + yz + zx) = 50, then find the value of (x² + y² + z²).
- If [4a - (1/2a)] = 6, then find the value of [16a² + (1/4a²) - 8]