The equation sqrt {3 {x^2} + x + 5} = x | vedclass

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Question

Consider $\sqrt{3 x^{2}+x+5}=x-3$

Squaring both the sides, we get

$3 x^{2}+x+5=(x-3)^{2}$

$\Rightarrow 3 x^{2}+x+5=x^{2}+9-6 x$

$\Rightarr

Vedclass · Accepted Answer

no solution

Vedclass · Answer

Consider $\sqrt{3 x^{2}+x+5}=x-3$

Squaring both the sides, we get

$3 x^{2}+x+5=(x-3)^{2}$

$\Rightarrow 3 x^{2}+x+5=x^{2}+9-6 x$

$\Rightarrow 2 x^{2}+7 x-4=0$

$\Rightarrow 2 x^{2}+8 x-x-4=0$

$\Rightarrow 2 x(x+4)-1(x+4)=0$

$\Rightarrow x=\frac{1}{2}$ or $x=-4$

For $x=\frac{1}{2}$ and $x=-4$

$L.H.S.$ $
eq$ $R.H.S.$ of equation, $\sqrt{3 x^{2}+x+5}$ $=x-3$

Also, for every $x \in R, \mathrm{LHS} 
eq \mathrm{RHS}$ of the given equation. 

$	herefore $ Given equation has no solution.

The equation $\sqrt {3 {x^2} + x + 5} = x - 3$ , where $x$ is real, has

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