HUMRAJ!!!!!!!!
If x, y and z are real numbers such that x +y +z = 5 and xy +yz + zx =3, then which of the following equations best descibes the largest value of x?
a. x ≤ 5/3
b. x < v19
c. x ≤ 13/3
d. x < 4
(x+y+z)²=x²+y²+z²+2(xy+yz+xz);
25=x²+y²+z² +2*3; x²=25-6-(y²+z² )=19-(y²+z² ); Now x will be maximum if (y²+z² ) =0 Hence answer is x<v19.
@ Randhir
If y2 + z2 = 0; then y and z both are zero.
and xy + yz + zx = 0 which is not possible.
Am I right???????????
x < v19 what do you mean by v
@ NILAY
Yaar, square root of 19.
the solution of answer s x < v19.....................and moreover (Y2 + Z2) WILL NOT equall to zero but t wll tend to zero and thus x wll tend to v19 thus x wll be maximum at x < v19....................
Let y=z
Then ,
the two eqns r simplified to
x+2y=5
2xy+y^2=3
Solving the 2 eqns we get
y=1/3 or 3
since we hv 2 find the max value of x...so y=z=1/3
theefore,
max value of x =13/3...
Regards,
Mayungk Kapur
n/a
therefore,
More information about formatting options
(x+y+z)²=x²+y²+z²+2(xy+yz+xz);
25=x²+y²+z² +2*3; x²=25-6-(y²+z² )=19-(y²+z² ); Now x will be maximum if (y²+z² ) =0 Hence answer is x<v19.
@ Randhir
If y2 + z2 = 0; then y and z both are zero.
and xy + yz + zx = 0 which is not possible.
Am I right???????????
x < v19
what do you mean by v
@ NILAY
Yaar, square root of 19.
the solution of answer s x < v19.....................and moreover (Y2 + Z2) WILL NOT equall to zero but t wll tend to zero and thus x wll tend to v19 thus x wll be maximum at x < v19....................
Let y=z
Then ,
the two eqns r simplified to
x+2y=5
2xy+y^2=3
Solving the 2 eqns we get
y=1/3 or 3
since we hv 2 find the max value of x...so y=z=1/3
theefore,
max value of x =13/3...
Regards,
Mayungk Kapur
n/a
Let y=z
Then ,
the two eqns r simplified to
x+2y=5
2xy+y^2=3
Solving the 2 eqns we get
y=1/3 or 3
since we hv 2 find the max value of x...so y=z=1/3
therefore,
max value of x =13/3...
Regards,
Mayungk Kapur
n/a
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