If x + y +z= 19, x2 +y2+z2 =133 and xz =
y2, then the difference between z and x is:
Answers
Answer:
Difference of 5 is there between z and x.
Step-by-step explanation:
Given,
x + y + z = 19 ...( 1 )
x^2 + y^2 + z^2 = 133 ...( 2 )
xz = y^2 ...( 3 )
Simplifying ( 3 ) :
= > zx = y^2
= > z × x = y × y
= > x / y = y / z ... ( 4 )
Or, y / x { that's reciprocal of x / y } = z / y
Dividing both sides of ( 1 ) by y :
= > ( x + y + z ) / y = 19 / y
= > x / y + y / y + z / y = 19 / y
= > x / y + 1 + z / y = 19 / y
= > x / y + z / y = 19 / y - 1
Let, x / y be a. So, z / y should be 1 / a, since it's equal to the reciprocal of x / y.
= > a + 1 / a = 19 / y - 1
= > ( a + 1 / a )^2 = 19 / y - 1
= > a^2 + 1 / a^2 + 2( a × 1 / a ) = ( 19 / y - 1 )^2
= > a^2 + 1 / a^2 + 2 = ( 19 / y )^2 + 1 - 2( 19 / y )
= > a^2 + 1 / a^2 = 361 / y^2 - 1 - 38 / y ... ( 5 )
Dividing both sides of ( 2 ) by y^2 :
= > ( x^2 + y^2 + z^2 ) / y^2 = 133 / y^2
= > x^2 / y^2 + y^2 / y^2 + z^2 / y^2 = 133 / y^2
= > x^2 / y^2 + 1 + z^2 / y^2 = 133 / y^2
= > a^2 + 1 + 1 / a^2 = 133 / y^2
= > a^2 + 1 / a^2 = 133 / y^2 - 1 ... ( 6 )
Comparing the value of a^2 + 1 / a^2 from ( 5 ) and ( 6 ) :
= > 361 / y^2 - 1 - 38 / y = 133 / y^2 - 1
= > 361 / y^2 - 133 / y^2 - 38 / y = 0
= > ( 361 - 133 - 38y ) / y^2 = 0
= > 228 - 38y = 0
= > 228 = 38y
= > 228 / 38 = y
= > 6 = y
Now,
= > z - x
= > ± √( z - x )^2
= > ± √( z^2 + x^2 - 2xz )
= > ± √( z^2 + x^2 + y^2 - y^2 - 2xz )
= > ± √[ 133 - y^2 - 2y^2 ] { from above, zx = y^2 }
= > ± √[ 133 - 3y^2 ]
= > ± √[ 133 - 3( 6 )^2 ]
= > ± √[ 133 - 3( 36 ) ]
= > ± √( 133 - 108 )
= > ± √( 25 )
= > ± 5
Hence, difference of 5 is there between z and x.
Answer:
5
Step-by-step explanation:
THIS IS EASY WAY TO SOLVE THIS QUESTION
