A biquadratic equation to have at least two real roots

Solution 1:

This is a case of a "hidden quadratic" in that your equations are all really quadratic equations in a different variable. To see this, write $u=x^2$. Then for example your first equation becomes $$u^2-5u-36=0.$$ The temptation is then to look at the discriminant of this equation and conclude from there. But a solution of this equation gives us a value of $u=x^2$, which will give two values of $x$ if $u>0$, one (repeated) value if $u=0$ and no values if $u<0$. So in fact we have to solve this quadratic for $u$ and go from there. In this case we have $$u=\frac{5\pm\sqrt{25+144}}{2}=\frac{5\pm 13}{2}=9,-4.$$ This gives two values of $u$, but the case $u=x^2=-4$ has no real solutions for $x$ so we discard it. The case $u=x^2=9$ has two solutions $\pm 3$ for $x$ so overall there are two solutions.

Really this is most of the way to solving the equations, although we only need to know the sign of the values of $u$, so for example we can see that $$u=\frac{5\pm\sqrt{25+144}}{2}$$ clearly gives one positive solution without explicitly calculating.

Solution 2:

$$ x^4 -13 x^2 + 36 = (x^2 + 6)^2 - 25 x^2 = (x^2 + 5x+6) (x^2 - 5x+6) $$ and both quadratic factors have real roots(positive discriminants).

$$ 4 x^4 -10 x^2 + 25 = (2x^2 + 5)^2 - 30 x^2 = (2x^2 + x \sqrt{30}+5) (2x^2 - x \sqrt{30}+5) $$ and both quadratic factors have negative discriminants. OR $$ 4 x^4 -10 x^2 + 25 = (2x^2 - 5)^2 +10 x^2 $$ is always positive