A biquadratic equation to have at least two real roots

Question

Which of the equations have at least two real roots? \begin{aligned} x^4-5x^2-36 & = 0 & (1) \\ x^4-13x^2+36 & = 0 & (2) \\ 4x^4-10x^2+25 & = 0 & (3) \end{aligned} I wasn't able to notice something clever, so I solved each of the equations. The first one has $2$ real roots, the second one $4$ real roots and the last one does not have real roots. I am pretty sure that the idea behind the problem wasn't solving each of the equations. What can we note to help us solve it faster?

Answer 1

Descartes' rule of signs says that:

• 1 pos and 1 neg real root for the first one.
• 2,0 pos and 2,0 neg for the second and third one.

Thus the first one you know already that it has two real roots. For the second and third, you can use the discrimant to see that the third has no real roots and the second will have 2 or 4.

EDIT: Since you're just interested in whether you have 2 or more roots, the discrimant being positive is enough (as saulspatz pointed out)

Answer 2

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.

Answer 3

$$ x^4 -13 x^2 + 36 = (x^2 + 6)^2 - 25 x^2 = (x^2 + 5x+6) (x^2 - 5x+6) $$ 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

Answer 4

There is a test that can be made for biquadratics without the need for square-roots that is akin to a test for the existence of real zeroes of a quadratic polynomial (there is an evident connection with the Rule of Signs that Gregory discusses). There, "completing the square" produces $$ u^2 \ + \ Bu \ + \ C \ \ = \ \ \left(u \ + \ \frac{B}{2} \right)^2 \ + \ \left(C \ - \ \frac{B^2}{4} \right) \ \ = \ \ 0 \ \ \Rightarrow \ \ \left(u \ + \ \frac{B}{2} \right)^2 \ \ = \ \ \frac{B^2}{4} \ - \ C \ \ , $$ which tells us that there are no real zeroes for $ \ u \ $ if $ \ \frac{B^2}{4} - C \ < 0 \ \Rightarrow \ B^2 \ < \ 4C \ \ , $ one if $ \ B^2 \ = \ 4C \ \ , $ and two if $ \ B^2 \ > \ 4C \ \ . $ This is equivalently to saying that a parabola with its vertex at $ \ x \ = \ h \ = \ -\frac{B}{2} \ , \ y \ = \ k \ = \ C \ - \ \frac{B^2}{4} \ $ does not intersect the $ \ x-$axis in this first case, is just tangent to it in the second, and has two $ \ x-$intercepts in the third case.

With the biquadratic $ \ x^4 \ + \ Bx^2 \ + \ C \ \ , $ which substitutes $ \ u = x^2 \ \ , $ now requires $ \ \left(x^2 \ + \ \frac{B}{2} \right)^2 \ > \ 0 \ $ for $ \ B > 0 \ $ and the polynomial cannot have a value less than $ \ \left( \frac{B}{2} \right)^2 + \frac{B^2}{4} - C \ = \ C \ . $ From this, we see that the biquadratic has no real zeroes for $ \ \mathbf{B > 0 \ , \ C > 0} \ \ $ and two real zeroes for $ \ \mathbf{B > 0 \ , \ C < 0} \ \ . $ In terms of factorization, the first of these corresponds to $$ (x^2 + \alpha^2)·(x^2 + \beta^2) \ \ = \ \ x^4 \ + \ (\alpha^2 + \beta^2)·x^2 \ + \ \alpha^2·\beta^2 \ \ , $$ while the second is one case for $$ (x^2 + \alpha^2)·(x^2 - \beta^2) \ \ = \ \ x^4 \ + \ (\alpha^2 - \beta^2)·x^2 \ - \ \alpha^2·\beta^2 \ \ , $$ with $ \ |\alpha| > |\beta| \ \ , \ \ \alpha \ $ being imaginary.

Coming to the cases pertinent to your equations, we have $ \ B < 0 \ \ , $ which permits $ \ \left(x^2 \ + \ \frac{B}{2} \right)^2 \ \ge \ 0 \ \ . $ The significance of this change is that the biquadratic curve has just one "turning point" for $ \ B > 0 \ \ , $ but three for $ \ B < 0 \ \ . $ The function $ \ f(x) \ = \ x^4 \ + \ Bx^2 \ + \ C \ \ $ has even symmetry and can be thought of as "fusing" the portion of the parabola $ \ y \ = \ x^2 \ + \ Bx \ + \ C \ $ for $ \ x \ge 0 \ $ with its "reflection" about the $ \ y-$axis. For $ \ B > 0 \ \ , $ the vertex of the parabola is at $ \ x \ = \ -\frac{|B|}{2} \ < \ 0 \ \ , $ so the biquadratic curve only has a turning point at its $ \ y-$intercept $ \ (0 \ , \ C ) \ \ . $ However, for $ \ B < 0 \ \ , $ the vertex lies at $ \ x \ = \ \frac{|B|}{2} \ > \ 0 \ \ , $ leading to the turning point at $ \ (0 \ , \ C ) \ \ $ and two others at $ \ x^2 \ = \ \frac{|B|}{2} \ \Rightarrow \ x \ = \ \pm \sqrt{ \frac{|B|}{2}} \ \ . $ The $ \ y-$coordinate for these points is then $$ \left(\frac{|B|}{2} \right)^2 \ + \ B·\left(\frac{|B|}{2} \right) \ + \ C \ \ = \ \ \frac{ B^2 }{4} \ - \ |B|·\left(\frac{|B|}{2} \right) \ + \ C $$ $$ = \ \ \frac{ B^2 }{4} \ - \ \frac{ B^2}{2} \ + \ C \ \ = \ \ C \ - \ \frac{ B^2 }{4} \ \ . $$ [The foregoing can be found more directly by using calculus.]

With this established, we can now sort the cases for $ \ B < 0 \ \ : $

• with the "high" turning point "below" the $ \ x-$axis $ \ [ \ \mathbf{C < 0} \ ] \ \ , $ the biquadratic curve has two $ \ x-$intercepts [the factorization is $ (x^2 + \alpha^2)·(x^2 - \beta^2) \ \ = \ \ x^4 \ - \ (\beta^2 - \alpha^2)·x^2 \ - \ \alpha^2·\beta^2 \ \ , $ with $ \ |\alpha| < |\beta| \ \ , \ \ \alpha \ $ imaginary] ;

• with the other turning points "above" the $ \ x-$axis, for which $ \ C \ - \ \frac{ B^2 }{4} \ > \ 0 \ \Rightarrow \ \mathbf{B^2 \ < \ 4C} \ \ , $ the curve does not intersect the $ \ x-$axis;

• for $ \ C \ - \ \frac{ B^2 }{4} \ < \ 0 \ < \ C \ \Rightarrow \ \mathbf{0 \ < \ 4C \ < \ B^2} \ \ , $ there are four $ \ x-$intercepts "surrounding" the three turning points [the corresponding factorization is $ \ (x^2 - \alpha^2)·(x^2 - \beta^2) \ \ = \ \ x^4 \ - \ (\alpha^2 + \beta^2)·x^2 \ + \ \alpha^2·\beta^2 \ \ . \ ] $

We see then that for your equation (1), $ \ B = -5 \ , \ C = -36 \ $ tells us that the equation has two real zeroes; its factorization is $ \ (x^2 - 9)·(x^2 + 4) \ \ . $ Equation (2) has $ \ B = -13 \ , \ C = 36 \ $ $\Rightarrow \ B^2 = 169 \ > \ 4C = 144 \ > \ 0 \ \ , $ hence there are four real zeroes; the expression factors as $ \ (x^2 - 9)·(x^2 - 4) \ \ . $ Finally, with equation (3), we "factor out" a $ \ 4 \ \ $ (which has no effect on the solutions) to obtain $ \ B = -\frac52 \ , \ C = \frac{25}{4} \ \Rightarrow \ B^2 = \frac{25}{4} \ < \ 4C = 25 \ \ , $ indicating that there are no real zeroes. (The factorization here is not very tidy.)

A table of the conditions for the number of real zeroes is presented below.

Answer 5

All you need to do is

let $x^2=a$

The above equations will become so

$a^4−5a^2−36=0$

$a^4−13a^2+36=0$

$4a^4−10a^2+25=0$

And now calculate 'D' that is $b^2-4ac$

and the usual rules apply but since we first assumed $x^2=a$,

Hence the roots will be twice as many ie

4 for D>0

2 for D=0