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प्रश्न
If a – b + c = 4, ab + bc – ac = 17, find the value of a2 + b2 + c2.
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उत्तर
Given: a – b + c = 4, ab + bc – ac = 17
Step-wise calculation:
1. Start with the expression (a – b + c)2:
(a – b + c)2 = a2 + b2 + c2 – 2ab + 2bc – 2ac
2. Substitute the given values:
(a – b + c)2 = 42
(a – b + c)2 = 16
a2 + b2 + c2 – 2(ab – bc + ac) = 16
3. Notice the given second equation (ab + bc – ac = 17).
Rearranging it to fit the middle terms: ab + bc – ac = 17.
So, –2(ab – bc + ac) = –2(ab – bc + ac).
We need carefully to rewrite middle terms:
a2 + b2 + c2 – 2ab + 2bc – 2ac = 16
Observe: 2ab + 2bc – 2ac = –2(ab – bc + ac)
If we let S = ab – bc + ac, then a2 + b2 + c2 – 2S = 16.
We are given: ab + bc – ac = 17.
But we need (ab – bc + ac) to substitute for (S).
4. Find (S = ab – bc + ac) in terms of the given expression.
Note: (ab + bc – ac) + (ab – bc + ac) = 2ab.
So, 17 + S = 2ab
⇒ S = 2ab – 17.
But we don’t know (ab) directly.
5. Alternatively, re-express (S) in terms of known quantities:
Compare (S) and the given S = ab – bc + ac
Given: T = ab + bc – ac = 17
Add (S + T): S + T = (ab – bc + ac) + (ab + bc – ac)
S + T = 2ab
S + 17 = 2ab
⇒ S = 2ab – 17
6. Let’s use the identity:
(a – b + c)2 = a2 + b2 + c2 – 2ab + 2bc – 2ac
We were given: (a – b + c)2 = 16
Substitute: a2 + b2 + c2 – 2(ab – bc + ac) = 16
Rewrite as: a2 + b2 + c2 – 2S = 16
7. Since S = 2ab – 17,
Substitute back: a2 + b2 + c2 – 2(2ab – 17) = 16
a2 + b2 + c2 – 4ab + 34 = 16
a2 + b2 + c2 – 4ab = 16 – 34
a2 + b2 + c2 – 4ab = –18
8. We need an expression to find (ab) or (a2 + b2 + c2).
For this, note the identity: (a – b)2 = a2 + b2 – 2ab
We can try to express (a2 + b2 + c2) in terms of (ab), but lacking (a + b + c) or (a – b + c) sums, solve for (a2 + b2 + c2):
Rewrite: a2 + b2 + c2 – 4ab = –18
Add and subtract (2bc – 2ac):
a2 + b2 + c2 + 2bc – 2ac – 4ab – 2bc + 2ac = –18
Group terms: (a2 + b2 + c2 + 2bc – 2ac) – 4ab – 2bc + 2ac = –18
Recall from step 1: a2 + b2 + c2 – 2ab + 2bc – 2ac = 16
Then, 16 – 2ab = –18
–2ab = –34
ab = 17
9. Substitute (ab = 17) back into the earlier equation:
a2 + b2 + c2 – 4(17) = –18
a2 + b2 + c2 – 68 = –18
a2 + b2 + c2 = 50
