Question

Given that a^2+6b=-14, b^2+8c+23=0 and c^2+4a-8=0 then the value of ab+bc+ca is

Answer 1

Given that ,
a² + 6b = -14
b² + 8c = - 23
c² + 4a = 8

add all given equations ,
a² + 6b + b² + 8c + c² + 4a = - 14 - 23 + 8 = -29
⇒(a² + 4a + 4) - 4 + (b² + 6b + 9) - 9 + (c² + 8c + 16) - 16 = -29
⇒(a + 2)² + (b + 3)² + (c + 4)² - 4 - 9 - 16 = -29
⇒(a + 2)² + (b + 3)² + (c + 4)² = 0
⇒a = -2 , b = -3 and c = -4

Hence, ab + bc + ca = (-2)(-3) + (-3)(-4) + (-4)(-2)
= 6 + 12 + 8 = 26

Answer 2

Answer:

26

Step-by-step explanation:

a2+6b=−14(E01)

b2+8c+23=0⟹b2+8c=−23(E02)

c2+4a−8=0⟹c2+4a=8(E03)

(E01)+(E02)+(E03)=a2+6b+b2+8c+c2+4a=−14−23+8  

⟹(a2+4a)+(b2+6b)+(c2+8c)=−29  

⟹(a+2)2−4+(b+3)2−9+(c+4)2−16=−29  

⟹(a+2)2+(b+3)2+(c+4)2−29=−29  

⟹(a+2)2+(b+3)2+(c+4)2=0(E04)

Since sum of individual square terms are zero, each of the term must equal zero  

⟹a=−2,b=−3,c=−4  

ab+bc+ca=((−2)(−3))+((−3)(−4))+((−4)(−2))  

=6+12+8=26  

Ans: 26