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

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

Question

(C) Given equation: $a^{2}+b^{2}+c^{2}=2(a-b-c)-3$Expand the right side: $a^{2}+b^{2}+c^{2}=2a-2b-2c-3$Rearrange the terms to one side: $a^{2}-2a+b^{2

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Given equation: $a^{2}+b^{2}+c^{2}=2(a-b-c)-3$Expand the right side: $a^{2}+b^{2}+c^{2}=2a-2b-2c-3$Rearrange the terms to one side: $a^{2}-2a+b^{2}+2b+c^{2}+2c+3=0$Split $3$ into $1+1+1$: $(a^{2}-2a+1)+(b^{2}+2b+1)+(c^{2}+2c+1)=0$Rewrite as perfect squares: $(a-1)^{2}+(b+1)^{2}+(c+1)^{2}=0$Since the sum of squares is zero,each term must be zero: $a-1=0, b+1=0, c+1=0$Solving these gives: $a=1, b=-1, c=-1$Calculate $(a-b+c)$: $1-(-1)+(-1) = 1+1-1 = 1$

If $a^{2}+b^{2}+c^{2}=2(a-b-c)-3,$ then the value of $(a-b+c)$ is

Similar Questions

What fraction must be subtracted from the sum of $\frac{1}{4}$ and $\frac{1}{6}$ to have an average of $\frac{1}{12}$ of all the three fractions?

If $a^{2}+b^{2}+c^{2}=ab+bc+ac,$ then the value of $\frac{a+c}{b}$ is:

$67.99 \%$ of $1401 - 13.99 \%$ of $1299 = ?$

If $x = \sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}$,then what is the value of $(x^{2} - x - 9)$?

The value of $999 \frac{995}{999} \times 999$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module