Solving for \(a + b + c\)
The problem provides the equation: \(a^2 + b^2 + c^2 + 3 = 2(a + b + c)\) We need to find the value of \((a + b + c)\).
Algebraic Manipulation
First, rearrange the equation to bring all terms to one side:
\(a^2 + b^2 + c^2 - 2a - 2b - 2c + 3 = 0\)
Group the terms involving \(a\), \(b\), and \(c\) separately:
\((a^2 - 2a) + (b^2 - 2b) + (c^2 - 2c) + 3 = 0\)
Completing the Square
To solve this, we use the technique of completing the square for each variable. The general form is \((x - k)^2 = x^2 - 2kx + k^2\). We need \(k^2\) terms.
- For \((a^2 - 2a)\), we need \(+1\).
- For \((b^2 - 2b)\), we need \(+1\).
- For \((c^2 - 2c)\), we need \(+1\).
We can rewrite the equation by adding and subtracting these required terms:
\((a^2 - 2a + 1) - 1 + (b^2 - 2b + 1) - 1 + (c^2 - 2c + 1) - 1 + 3 = 0\)
Combine the perfect square terms and the constants:
\((a - 1)^2 + (b - 1)^2 + (c - 1)^2 - 1 - 1 - 1 + 3 = 0\)
\((a - 1)^2 + (b - 1)^2 + (c - 1)^2 + 0 = 0\)
This simplifies to:
\((a - 1)^2 + (b - 1)^2 + (c - 1)^2 = 0\)
Finding the Values of a, b, and c
The sum of squares of real numbers can only be zero if each individual term is zero. Therefore:
- \((a - 1)^2 = 0 \implies a - 1 = 0 \implies a = 1\)
- \((b - 1)^2 = 0 \implies b - 1 = 0 \implies b = 1\)
- \((c - 1)^2 = 0 \implies c - 1 = 0 \implies c = 1\)
Calculating the Final Value
Now, calculate the value of \((a + b + c)\):
\((a + b + c) = 1 + 1 + 1 = 3\)
The value of \((a + b + c)\) is 3.
Similar Questions
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- If $x + \frac{1}{x} = 5$
Then the value of $\frac{3x}{2x^2 + 2 - 5x}$ will be: - If $3a + 4b = 2$ and $ab = \frac{1}{36}$, then $27a^3 + 64b^3$ is:
Important Questions from Algebra
If 2x – y = 2 and xy = \(\frac{3}{2}\) , then what is the value of x 3– \(\frac{{{y^3}}}{8}\) ?
If (10a 3+ 4b 3) : (11a 3- 15b 3) = 7 : 5, then (3a + 5b) : (9a - 2b) =?
- If 4sin 2 θ = 3(1+ cos θ), 0° < θ < 90°, then what is the value of (2tan θ + 4sin θ - sec θ)?
The value of:
\(\frac{{\sin 23^\circ \cos 67^\circ + \sec52^\circ \sin38^\circ + \cos 23^\circ \sin 67^\circ + \rm cosec52^\circ \cos 38^\circ }}{{\rm cose{c^2}20^\circ - {{\tan }^2}70^\circ }}\)
If (x + y) 3+ 27(x - y) 3= (Ax - 2y)(Bx 2+ Cxy + 13y 2), then the value of A - B - C is:
