Question : If $a+b+c=0$, then the value of $\frac{a^{2}+b^{2}+c^{2}}{ab+bc+ca}$ is:
Option 1: 2
Option 2: –2
Option 3: 0
Option 4: 4
Correct Answer: –2
Solution : Given: $a+b+c=0$ We know that $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca)$ Putting the value of $a+b+c=0$, we have, ⇒ $(0)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca)$ ⇒ $a^{2}+b^{2}+c^{2}=–2(ab+bc+ca)$ ⇒ $\frac{(a^{2}+b^{2}+c^{2})}{(ab+bc+ca)}=–2$ Hence, the correct answer is –2.
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