Question

if a+b+c =0 , a²+b²+c² =k(a²-bc) find k​

Answer 1

Answer:

Step-by-step explanation:

If a + b + c = 0,

(a+b+c)² = a²+b²+c²+2ab+2ac+2bc

0² = a²+b²+c²+2ab+2ac+2bc

a²+b²+c²= -2ab-2ac-2bc

a²+b²+c²= -2(ab+ac+bc) ------1

If a + b + c = 0,

Then,

a = - b - c

b = - a - c

c = - b - a

a² - bc = a² - (- a - c )(- a - b)

= a² - (-a*-a + -c*-a + -c*-b + -a*-b)

= a² - (a² + ca + cb + ab)

= a² - a² - ca - cb - ab

= -1(ab + bc + ac)

Therefore a²-bc =-1(ab + bc + ac)------2

a²+ b² + c² = k(a² - bc)

k = (a²+ b² + c²)/(a² - bc)

Using ------2 and -----1, we get,

k = -2(a²+ b² + c²)/-1(a²+ b² + c²)

= -2/-1 = 2

Hope this helps.... :-)