Question:
Given that a, b, and c are real numbers satisfying the equations:
\(2a - 5b + 11c = 0\)
\(11a + 10b - 2c = 5\)
Find the value of the expression \( (a^2 - b^2 + c^2) \).
Given that a, b, and c are real numbers satisfying the equations:
\(2a - 5b + 11c = 0\)
\(11a + 10b - 2c = 5\)
Find the value of the expression \( (a^2 - b^2 + c^2) \).
\(2a - 5b + 11c = 0\)
\(11a + 10b - 2c = 5\)
Find the value of the expression \( (a^2 - b^2 + c^2) \).
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When the number of variables is greater than the number of linear equations, you cannot find a unique solution for the variables. However, a specific expression involving them might be constant. Squaring and adding/subtracting the equations is a powerful technique to try.
Updated On: Jun 30, 2026
- 5/11
- 2/13
- 1
- CBD (Cannot be determined)
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The Correct Option is D
Solution and Explanation
Approach: With two equations and three unknowns, isolate the symmetric combination by squaring-and-adding so the cross term \(ac\) cancels. Then read off honestly whether the target is pinned down.
Step 1: Rearrange. From \(2a - 5b + 11c = 0\): \(2a + 11c = 5b\). From \(11a + 10b - 2c = 5\): \(11a - 2c = 5 - 10b\).
Step 2: Square both and add. The cross terms are \(+44ac\) and \(-44ac\), which cancel: \[ (2a+11c)^2 + (11a-2c)^2 = 125a^2 + 125c^2. \] On the right, \((5b)^2 + (5-10b)^2 = 25b^2 + 25 - 100b + 100b^2 = 125b^2 - 100b + 25.\)
Step 3: Equate and divide by \(25\): \(5a^2 + 5c^2 = 5b^2 - 4b + 1\), so \[ 5(a^2 - b^2 + c^2) = 1 - 4b. \]
Step 4 (the honest read): The right side still contains \(b\). Since the system has \(3\) unknowns and only \(2\) equations, \(b\) is free, so \(a^2 - b^2 + c^2\) is not fixed. Check: with \(b=0\), a valid solution \(a = \tfrac{11}{25},\, c = -\tfrac{2}{25}\) gives \(a^2 - b^2 + c^2 = \tfrac{1}{5}\); with \(b=-1\) it gives \(1\). Two different values are possible.
Answer: The expression cannot be determined \(\Rightarrow \boxed{\text{CBD (option 4)}}\). (Note: the stored key marks option 3 = 1; this appears to be an error and is flagged.)
Step 1: Rearrange. From \(2a - 5b + 11c = 0\): \(2a + 11c = 5b\). From \(11a + 10b - 2c = 5\): \(11a - 2c = 5 - 10b\).
Step 2: Square both and add. The cross terms are \(+44ac\) and \(-44ac\), which cancel: \[ (2a+11c)^2 + (11a-2c)^2 = 125a^2 + 125c^2. \] On the right, \((5b)^2 + (5-10b)^2 = 25b^2 + 25 - 100b + 100b^2 = 125b^2 - 100b + 25.\)
Step 3: Equate and divide by \(25\): \(5a^2 + 5c^2 = 5b^2 - 4b + 1\), so \[ 5(a^2 - b^2 + c^2) = 1 - 4b. \]
Step 4 (the honest read): The right side still contains \(b\). Since the system has \(3\) unknowns and only \(2\) equations, \(b\) is free, so \(a^2 - b^2 + c^2\) is not fixed. Check: with \(b=0\), a valid solution \(a = \tfrac{11}{25},\, c = -\tfrac{2}{25}\) gives \(a^2 - b^2 + c^2 = \tfrac{1}{5}\); with \(b=-1\) it gives \(1\). Two different values are possible.
Answer: The expression cannot be determined \(\Rightarrow \boxed{\text{CBD (option 4)}}\). (Note: the stored key marks option 3 = 1; this appears to be an error and is flagged.)
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