Question

If $x=1+a^2,y=1+b^2,z=1+c^2$ and $(a+b+c{)}^2=0$, then ab + bc + ca =

• A. [3 – (x + y + z)]/2
• B. 1 – (x + y + z)/2
• C. 1 + (x + y + z)/2
• D. 1 – (x + y + z)

Answer 1

The correct option is A [3 – (x + y + z)]/2

a = 0, b = 1, c = -1. x = 1, y = 2, c = 2. ab + bc + ca = -1. Only option a) satisfies

Alternatively: -

Given

$x=1+a^2,\Rightarrow a^2=x-1$

$y=1+b^2,\Rightarrow b^2=y-1$

$z=1+c^2,\Rightarrow c^2=z-1$ and $(a+b+c{)}^2=0$

As we know, $(a+b+c{)}^2=a^2+b^2+c^2+2(ab+bc+ca)$

Put the given values.

0 = (x - 1) + (y - 1) + (z - 1) + 2(ab + bc + ca)

(ab + bc + ca) = $\frac{3-(x+y+z)}{2}$