Question : The value of $x$ which satisfies the equation $\frac{x \:+\: a^2 \:+\: 2c^2}{b \:+\: c} + \frac{x \:+\: b^2 \:+\: 2a^2}{c+a} + \frac{x \:+\: c^2 \:+\: 2b^2}{a \:+\: b} = 0$ is:
Option 1: $(a^2+b^2+c^2)$
Option 2: $- (a^2+b^2+c^2)$
Option 3: $(a^2+2b^2+c^2)$
Option 4: $-(a^2+b^2+2c^2)$
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
Correct Answer: $- (a^2+b^2+c^2)$
Solution :
Given: $\frac{x+a^2+2c^2}{b+c}+\frac{x+b^2+2a^2}{c+a}+\frac{x+c^2+2b^2}{a+b}=0$
We know the algebraic identity, $(x^2-y^2)=(x+y)(x-y)$.
Substitute the value of $x=-(a^2+b^2+c^2)$ in the given equation, we get,
$\frac{–a^2–b^2–c^2+a^2+2c^2}{b+c}+\frac{–a^2–b^2–c^2+b^2+2a^2}{c+a}+\frac{–a^2–b^2–c^2+c^2+2b^2}{a+b}=0$
$⇒\frac{c^2–b^2}{b+c}+\frac{a^2–c^2}{c+a}+\frac{b^2–a^2}{a+b}=0$
$⇒\frac{(c–b)(b+c)}{b+c}+\frac{(a–c)(c+a)}{c+a}+\frac{(b–a)(a+b)}{a+b}=0$
$⇒c-b+a-c+b-a=0$
$⇒0=0$
It satisfies the equation.
Hence, the correct answer is $-(a^2+b^2+c^2)$.
Related Questions
Know More about
Staff Selection Commission Combined Grad ...
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Get Updates Brochure
Your Staff Selection Commission Combined Graduate Level Exam brochure has been successfully mailed to your registered email id “”.
