The number of solutions of $\sqrt {3x^2+x+5}=x-3$ is

Question

Squaring on both sides we get the equation $$2x^2+7x-4=0$$ Then $$x=\frac 12$$ and $$x=-4$$ They do indeed satisfy the original equation but aren’t a part of the answer. I get that it must have to be the $\pm \sqrt {a}$ but I would still like a thorough explanation

Thanks!

Answer 1

They do not in fact satisfy the original equation.

For $x=\frac{1}{2}$, you get $\frac{5}{2}=-\frac{5}{2}$ and $x=4$, you get $7=-7$. Both are clearly false.

A common misconception is assuming that $\sqrt{a}$ assumes two values, $\pm\sqrt{a}$ simultaneously because it is the 'inverse' of the square function. However, this is not true, and we only take the positive branch of the square root function. This is because we want a function to be well-defined, to map to a unique number, instead of 2.

Answer 2

Hint

As $x-3=\sqrt{3x^2+x+5}\ge0$ for real $x$

$x\ge3$

Answer 3

Solving the equation $$2x^2+7x-4=0$$ we get $$x^2+\frac{7}{2}x-2=0$$ (dividing by $2$): and by the quadratic formula $$x_{1,2}=-\frac{7}{4}\pm\sqrt{\frac{49}{16}+2}$$ $$x_{1,2}=-\frac{7}{4}\pm\frac{9}{4}$$ we get $$x_1=\frac{1}{2}$$ or $$x_2=-4$$ but for the squaring we need the condition $$x\geq 3$$ so we get no solution.