If a+b+c=0 and ab+bc+ca=25 find a
Answer
Given,
a+b+c=10 and ab+bc+ca=25
We know,
(a+b+c)2=(a2+b2+c2)+2(ab+bc+ca)
Thus,
(a+b+c)2=(a2+b2+c2)+2(ab+bc+ca)
=>102=a2+b2+c2+2(25)
=>100−50=a2+b2+c2
=>a2+b2+c2=50
Given,
a+b+c=10 and ab+bc+ca=25
We know,
(a+b+c)2=(a2+b2+c2)+2(ab+bc+ca)
Thus,
(a+b+c)2=(a2+b2+c2)+2(ab+bc+ca)
=>102=a2+b2+c2+2(25)
=>100−50=a2+b2+c2
=>a2+b2+c2=50
Answers
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Answer:
Given,
a+b+c=10 and ab+bc+ca=25
We know,
(a+b+c)2=(a2+b2+c2)+2(ab+bc+ca)
Thus,
(a+b+c)2=(a2+b2+c2)+2(ab+bc+ca)
=>102=a2+b2+c2+2(25)
=>100−50=a2+b2+c2
=>a2+b2+c2=50
Step-by-step explanation:
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Given,
a+b+c=10 and ab+bc+ca=25
We know,
(a+b+c)
2
=(a
2
+b
2
+c
2
)+2(ab+bc+ca)
Thus,
(a+b+c)
2
=(a
2
+b
2
+c
2
)+2(ab+bc+ca)
=>10
2
=a
2
+b
2
+c
2
+2(25)
=>100−50=a
2
+b
2
+c
2
=>a
2
+b
2
+c
2
=50