Solve the the given equation: $\sqrt{3x^2+x+5} = x-3$

Question

We have to find the number of solution for the given equation: $$\sqrt{3x^2+x+5} = x-3.$$ There are two solution one is

By using graph we get one solution

By squaring both sides we get no solution

I want to know which solution is correct

Answer 1

Hint:

The given equation is equivalent to the system: $$ \begin{cases} 3x^2+x+5=(x-3)^2\\ x-3\ge 0 \end{cases} $$

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Note that the system has no solutions because the roots of the second degree equations : $$ x=\frac{-7\pm\sqrt{79}}{4} $$ are less than $3$.

And this is in accord with the fact that the graphs of the two functions $$ y=\sqrt{3x^2+x+5} \qquad y=x-3 $$ have no common points (in your graph you have the wrong function $x=3$).

Answer 2

we have $$\sqrt{3x^2+x+5}=x-3$$ it must be $$x\geq 3$$ and after squaring we get $$3x^2+x+5=x^2-6x+9$$ this is equivalent to $$2x^2+7x-4=0$$ solving this quadratic equation we obtain $$x_{1,2}=-\frac{7}{4}\pm\sqrt{\frac{49}{16}+\frac{32}{16}}$$ Can you finish this?

Answer 3

x=-4 3*16-4+5=49 (-3-4)(-3-4)=49 So this equation was wrong first place.