Let a, b, c be non-zero real numbers such tha | vedclass

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

Question

(b)

Given, $a+b+c=0 \Rightarrow a, b, c \in R$

$\Rightarrow \quad a^2+b^2+c^2=q$

$\Rightarrow \quad a^4+b^4+c^4=r$

$\Rightarrow \quad\left

Vedclass · Accepted Answer

$q^2=2 r$ always

Vedclass · Answer

(b)

Given, $a+b+c=0 \Rightarrow a, b, c \in R$

$\Rightarrow \quad a^2+b^2+c^2=q$

$\Rightarrow \quad a^4+b^4+c^4=r$

$\Rightarrow \quad\left(a^2+b^2+c^2ight)^2=a^4+b^4+c^4 +2\left(a^2 b^2+b^2 c^2+c^2 a^2ight)$

$\left(a^2+b^2+c^2ight)^2=a^4+b^4+c^4+2 \left[(a b+b c+c a)^2-2 a b c(a+b+c)ight]$

$q^2=r+2\left[(a b+b c+c a)^2-2(a b c)(0)ight]$

$q^2=r+2[a b+b c+c a]^2$

$q^2=r+2  \left[\begin{array}{c}(a+b+c)^2-\left(a^2+b^2+c^2ight) \ 2\end{array}ight]^2$

$q^2=r+2\left(\frac{0-q}{2}ight)^2$

$q^2=r+\frac{2 q^2}{4}$

$q^2-\frac{1}{2} q^2=r$

$q^2=2 r$

Let $a, b, c$ be non-zero real numbers such that $a+b+c=0$, let $q=a^2+b^2+c^2$ and $r=a^4+b^4+c^4$. Then,

Similar Questions

If the roots of the equation $8{x^3} - 14{x^2} + 7x - 1 = 0$ are in $G.P.$, then the roots are

If $\alpha, \beta$ are the roots of the equation, $x^2-x-1=0$ and $S_n=2023 \alpha^n+2024 \beta^n$, then

The sum of the cubes of all the roots of the equation $x^{4}-3 x^{3}-2 x^{2}+3 x+1=10$ is

Consider the equation $(1+a+b)^2=3\left(1+a^2+b^{2})\right.$ where $a, b$ are real numbers. Then,

The number of cubic polynomials $P(x)$ satisfying $P(1)=2, P(2)=4, P(3)=6, P(4)=8$ is