If $a^{2} + b^{2} + 2 c^{2} - 4 a + 2 c - 2 a b + 5 = 0$ then the possible value of $a + b - c$ is:

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

- 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $2$
Consider the given expression $a^{2} + b^{2} + 2 c^{2} - 4 a + 2 c - 2 b c + 5 = 0$ and split it as follows:

$a^{2} + b^{2} + c^{2} + c^{2} - 4 a + 2 c - 2 b c + 4 + 1 = 0$

Now, combine the terms for completing squares as shown below:

$a^{2} + b^{2} + c^{2} + c^{2} - 4 a + 2 c - 2 b c + 4 + 1 = 0 \Rightarrow (a^{2} + 4 - 4 a) + (b^{2} + c^{2} - 2 b c) + (c^{2} + 1 + 2 c) = 0 \Rightarrow (a - 2)^{2} + (b - c)^{2} + (c + 1)^{2} = 0 . . . . . . . . (1)$

Equation 1 implies:

$(a - 2)^{2} = 0 \Rightarrow a - 2 = 0 \Rightarrow a = 2$

$(b - c)^{2} = 0 \Rightarrow b - c = 0 \Rightarrow b = c$

$(c + 1)^{2} = 0 \Rightarrow c + 1 = 0 \Rightarrow c = - 1 = b$

Therefore, $a = 2, b = - 1$ and $c = - 1$

Now, $a + b - c = 2 - 1 - (- 1) = 2 - 1 + 1 = 2$

Hence, the value of $a + b - c = 2$ .

Suggest Corrections

Similar questions

If a,b and c are real number such that a²+b²⁺2c²=4a-2c+2bc-5,then find the possible value of (6a-4b-2c).

If a + b = 5 and a - b = 1 find the value of a² + b²/2ab

a^{2} + b^{2} + c^{2} - a b - b c - c a \leq 0 \forall a, b, c \in R

∣ ∣ ∣ ∣ ∣ \begin{matrix} (a - b + 2)^{2} & a^{2} - b^{2} & 1 1 & (b - c + 2)^{2} & b^{2} - c^{2} c^{2} - a^{2} & 1 & (c - a + 2)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

If (a-b) sin ( $θ + ϕ$ ) = (a+b) sin ( $θ + ϕ$ ) and a tan $(\frac{θ}{2})$ - btan $(\frac{ϕ}{2})$ = C,the the value of sin $ϕ$ is equal

A = (\begin{matrix} 1 & - 4 - 2 & 3 \end{matrix})

B = (\begin{matrix} - 1 & 6 3 & - 2 \end{matrix})

(A + B)^{2} \neq A^{2} + 2 A B + B^{2}

Join BYJU'S Learning Program