Question

If a+b+c=4 and ab+bc+ca =4 then a²+b²+c²​

Answer 1

$(a + b + c) { }^{2} = a {}^{2} + b {}^{2} + c {}^{2} + 2(ab + bc + ca) \\ (4 {}^{2} ) = a {}^{2} + b {}^{2} + c {}^{2} + 2(4) \\ 16 = a {}^{2} + b {}^{2} +c {}^{2} + 8 \\ 16 - 8 = a {}^{2} + b {}^{2} + c {}^{2} \\ 8 = a {}^{2} + b {}^{2} + c {}^{2}$

hope it helps.......

Answer 2

We know that,

(a+b+c)² = a²+b²+c²+2ab+2bc+2ca

Let’s take

x = (a+b+c)²

y = a²+b²+c²

z = ab+bc+ca

therefore,

x = y+2z……………………….. (1)

by given condition,

5 (a²+b²+c²) = 4 (ab+bc+ca)

5y = 4z………………………… (2)

We have 2 eqns and 3 unknowns,

(1)…… x - y - 2z = 0

(2)….. 0x + 5y – 4z = 0

Using the formula,

Let’s say,

Then,

(x, y, z) = (14k, 4k, 5k) where k = 0, 1, 2, 3 ……. N

Therefore,

x = 14k

(a+b+c)² = 14k

(a+b+c) = √14k, where k = 0, 1, 2, 3 ……. n.