Question

If x^2 - 2y = -13, y^2 - 4z =14 , z^2+ 6x = -15 then find the value of xy + xz + 6yz

Answer 1

xy + xz + 6yz = 3 if x² - 2y = - 13 , y² - 4z = 14 , z² + 6x = = -15

Step-by-step explanation:

x² - 2y = - 13

y² - 4z = 14

z² + 6x = = -15

x² + y² + z²  + 6x - 2y - 4z  = - 14

(x + 3)² - 9 + (y - 1)²  - 1   + (z - 2)² - 4 = -14

=> (x + 3)²   + (y - 1)²  + (z - 2)² = 0

=> x = - 3  , y = 1  , z = 2

xy + xz + 6yz

= (-3)*1  +(-3)*2 + 6*1*2

= -3 - 6 + 12

= 3

xy + xz + 6yz = 3

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Answer 2

$x^{2} - 2y = - 13...........(1)\\y^{2} - 4z = 14..............(2)\\z^{2} + 6x = - 15...........(3)</p><p>$

Adding Equations (1) , (2) & (3) -

We have ,

x²+y²+z²-2y-4z+6x = -13+14-15

=>x²+6x+y²-2y+z²-4z = -14

=>(x²+2×x×3+3²)-9+(y²-2×y×1+1²)-1+(z²-2×z×2+2²)-4=-14

=>(x+3)²+(y-1)²+(z-2)²-9-1-4 = -14

=>(x+3)²+(y-1)²+(z-2)² - 14 = -14

=>(x+3)²+(y-1)²+(z-2)² = -14 + 14

=>(x+3)²+(y-1)²+(z-2)² = 0

If adding (3 distinct numbers) -

A + B + C = 0

Then ,

A = 0

B = 0

C = 0

So,

(x+3)² = 0

(y-1)² = 0

(z-2)² = 0

Then ,

x = -3

y = 1

z = 2

Now ,

xy + xz + 6yz (Putting the values of x,y & z)

(-3×1) + (-3×2) + (6×1×2)

= -3 + (-6) + 12

= -3 - 6 + 12

= -9 + 12

= 0

My ANSWER is ABSOLUTELY Correct.....My GUARANTEE.....Am I RIGHT or Am I RIGHT