If 3(a^{2}+b^{2}+c^{2})=(a+b+c)^{2}, then the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given equation: $3(a^{2}+b^{2}+c^{2})=(a+b+c)^{2}$Expanding the right side: $3a^{2}+3b^{2}+3c^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ac$Rearranging the te

Vedclass · Accepted Answer

$a=b=c$

Vedclass · Answer

(C) Given equation: $3(a^{2}+b^{2}+c^{2})=(a+b+c)^{2}$Expanding the right side: $3a^{2}+3b^{2}+3c^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ac$Rearranging the terms: $2a^{2}+2b^{2}+2c^{2}-2ab-2bc-2ac=0$Dividing by $2$: $a^{2}+b^{2}+c^{2}-ab-bc-ac=0$Multiplying by $2$: $2a^{2}+2b^{2}+2c^{2}-2ab-2bc-2ac=0$This can be rewritten as: $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0$Since the sum of squares is zero,each term must be zero:$(a-b)^{2}=0 \Rightarrow a=b$$(b-c)^{2}=0 \Rightarrow b=c$$(c-a)^{2}=0 \Rightarrow c=a$Therefore,$a=b=c$.

If $3(a^{2}+b^{2}+c^{2})=(a+b+c)^{2},$ then the relation between $a, b$ and $c$ is

Similar Questions

Let one root of $ax^2 + bx + c = 0$,where $a, b, c$ are integers,be $3 + \sqrt{5}$. Then the other root is:

The number of real roots of the equation $\frac{P^2}{x} + \frac{Q^2}{x - 1} = 1$,where $P$ and $Q$ are non-zero real numbers,is

If $\alpha$ and $\beta$ are the roots of the equation $4x^2 + 3x + 7 = 0$,then $\frac{1}{\alpha} + \frac{1}{\beta} =$

In the following,determine the set of values of $P$ for which the given quadratic equation $Px^2 + 4x + 1 = 0$ has real roots.

Vedclass Test Series

Exam Paper Generator

Online Exam Module