Quadratic discriminant in a 4th degree equation?

The existence of at least one real solution for the given equation implies $a^2-4b\geq 0$ or $a<-1.$ Let us prove the contrapositive of it. We are going to prove that if $a^2-4b< 0$ and $a\geq -1$ then the given equation does not have real solutions. We can rewrite the given polynomial equation in the form $x^2=\frac{x-1}{x^2+ax+b}.\:\:\:\:\:\:\:(*)$ Since $a^2-4b<0$ the domain of the rational function in the right side is the set of all real numbers. We can also see that the denominator is always positive and therefore the sign of it depends on the sign of $x-1$ , therefore, its sign is negative for $x<1$ and positive for $x>1,$ and, of course it is zero at zero. Since $x^2$ is always nonnegative then the equation $(*)$ does not have any real solution less than or equal to 1. Now, let us prove that the equation does not have solution greater than 1. The derivative of $f(x)=\frac{x-1}{x^2+ax+b}$ is $f '(x)= \frac{-x^2+2 x+a+b}{(x^2+a x+b)^2}$ that has two zeros and is not undefined at any number. Discussing the sign of $f '(x)$ we obtain that the function $f(x)$ has a local maximum on the interval $(1, \infty)$ at the $x = \sqrt{a+b+1}+1$ . The local maximum value is $\frac{1}{a+2 \sqrt{a+b+1}+2}$ . Using that $a^2<4b$ we obtain that $a+b+1>a+\frac{a^2}{4}+1=(\frac{a}{2}+1)^2\geq 0,$ so the radical in the given expression is a real number. Besides that, using that $a\geq -1$ we can obtain the following: $\frac{1}{a+2 \sqrt{a+b+1}+2}<\frac{1}{a+2+2\sqrt{a+a^2/4+1}}=\frac{1}{4+2a}<\frac{1}{2}.$ Since the function $x^2$ in the left side of $(*)$ has a minimum value of 1 on the interval $[1, \infty)$ and the function $\frac{x-1}{x^2+ax+b}.$ has maximum less than $\frac{1}{2}$ on the same interval, their graphs never intersect. Therefore $(*)$ does not have any real solution on $(1, \infty).$ This proves our claim.

Now let us disprove the other options.

Disproving the condition $a^2-4b\geq 0$ and $a<-1.$

Just use the counterexample: $x^4-\frac{1}{2}x^3-\frac{1}{2}x^2-x+1=0$ .

The previous equation has at least a real zero, namely $x=1.$ In this case, $a=-\frac{1}{2}$ and $b=-\frac{1}{2}.$ So $a^2-4b\geq 0$ but $a>-1.$

Disproving the condition $a^2-4b< 0$

Use the previous counterexample also.

Disproving the condition $a^2-4b\geq 0$

Use the counterexample $x^4-1.99x^3+0.9999x^2-x+1,$ where $a=-1.99$ and $b=.9999.$ The condition $a^2-4b\geq 0$ would not be true in this case, but we can see that the equation has to have a solution on the interval $[1, 1.2].$ Evaluate the given polynomial at the end points of this interval to see that a change of sign occurs. Q.E.D.