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

Let $\mathrm{S}$ be the set of positive integral values of $a$ for which $\frac{\mathrm{ax}^2+2(\mathrm{a}+1) \mathrm{x}+9 \mathrm{a}+4}{\mathrm{x}^2-8 \mathrm{x}+32}<0, \forall \mathrm{x} \in \mathbb{R}$. Then, the number of elements in $\mathrm{S}$ is :

The number of distinct real roots of the equation $x^{5}\left(x^{3}-x^{2}-x+1\right)+x\left(3 x^{3}-4 x^{2}-2 x+4\right)-1=0$ is

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

The number of solution$(s)$ of the equation $2^x = x^2$ is

Let $\mathrm{S}=\left\{x \in R:(\sqrt{3}+\sqrt{2})^x+(\sqrt{3}-\sqrt{2})^x=10\right\}$. Then the number of elements in $\mathrm{S}$ is :