I actually stumbled upon this question as a misprint in a Mathematics Problem book by Prilepko.
The original question was to find the values of $a$ for which the equation $x^2+2a\sqrt{a^2-3x}+4=0$ will have equal roots. The misprint was in the placement of $x$, which had to be outside the square root and then, by putting the discriminant of the quadratic equation to be zero, I could find the values of $a$ and the answer matched.
But I was wondering whether it was possible to solve the original question (i.e. without the misprint). Let's say the task was to find the values of $a$ for which the equation above had, say, –
(i) 2 equal roots
(ii) 4 equal roots
Would there be any way to do this problem? I would be glad to know if any numerical technique could be used in this case, although all other types of answers are also welcome.
PS – Even if there would be no such value of $a$ where we can have equal roots, I would like to have a proof/argument for the same, if possible.
Best Answer
So, if I understand correctly, you already know the solution for $x^2+2ax\sqrt{a^2-3}+4=0$, and you want to solve $$x^2+2a\sqrt{a^2-3x}+4=0$$ Then first get rid of the square root $$2a\sqrt{a^2-3x}=-4-x^2$$ $$4a^2(a^2-3x)=(-4-x^2)^2$$ $$4a^2(a^2-3x)-(-4-x^2)^2=0$$ This is a quartic polynomial, so calculating its roots is difficult, but to know if at least two of its roots are equal all we need is its discriminant (this works for polynomials of arbitrary degree) $$-16384 a^{12}-1157888 a^8+2359296 a^4=0$$ this is a third-degree polynomial in $a^4$, which we can solve $$a^4=0$$ or $$a^4=\frac{1}{128} \left(-4523\pm 485 \sqrt{97}\right)$$ $a^4=0$ is not a valid solution for the original equation, but $a=\left(\frac{1}{128} \left(-4523-485 \sqrt{97}\right)\right)^{\frac{1}{4}}$ does make two (imaginary) roots equal, and $a=-\left(\frac{1}{128} \left(-4523+485 \sqrt{97}\right)\right)^{\frac{1}{4}}$ makes it have two equal real roots.