Question

If x + y +z= 19, x2 +y2+z2 =133 and xz =
y2, then the difference between z and x is:​

Answer 1

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 2

Answer:

5

Step-by-step explanation:

THIS IS EASY WAY TO SOLVE THIS QUESTION