Question

Find out the value of a, b and c when 2(ab+bc+ca)=192 , abc=144 , (a^2+b^2+c^2)=169

Answer 1

Answer:

(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca

(a+b+c )^2=a^2+b^2+c^2+2

(ab+bc+ca)

putting the values

144^2=a^2+b^2+c^2+192

20736=a^2+b^2+c^2+192

a^2+b^2+c^2=20736-192=20644

Answer 2

The value of a = 3 , b = 4 and c = 12

Explanation:

Given that,

$2(ab+bc+ca)=192$

$abc=144$

$a^2+b^2+c^2=169$

Using identity,

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$

$(a+b+c)^2=169+192$

$(a+b+c)^2=361$

$a+b+c=19$

$ab+bc+ca=96$

$abc=144$

Let a, b and c be root of a cubic polynomial.

$x^3-(\text{sum of root})x^2+(\text{sum of product of two zeros})x-\text{product of zeros}=0$

$x^3-(a+b+c)x^2+(ab+bc+ca)x-abc=0$

Substitute the values

$x^3-19x^2+96x-144=0$

solve above equation

Zeros are:

a = 3

b = 4

c = 12

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