Haunting quadratic

Calculus Level 4

$\large \left(e^{\frac{a}{2}}+e^{-\frac{a}{2}}\right)x^2 + 2ax + e^{\frac{a}{2}}+e^{-\frac{a}{2}}=0, \quad a\in \mathbb{R}$

What can you say about the roots of above quadratic?

Note: A purely imaginary number is a complex number whose real part is 0.

Nature of roots depend on the values of

a

. Both roots are purely imaginary. Can't be determined. Both roots are either purely imaginary, or real and equal. Both roots must be non-real. Both roots are real and distinct.

1 solution

Jason Martin
Nov 12, 2017

We can simplify the equation to

$x^2+\frac{a}{cosh(a/2)}x+1=0$ , where $cosh(u)=\frac{e^u+e^{-u}}{2}$ . This has solutions

$x=-\frac{a/2}{cosh(a/2)} \pm \sqrt{\left(\frac{a/2}{cosh(a/2)}\right)^2-1}$ .

Now since $|b|<cosh(b)$ for all real $b$ , we know the discriminant is strictly negative. Thus, the roots are non-real.

Note: We can have imaginary roots, but only for $a=0$ . Thus, "non-real roots" is the best answer.

Haunting quadratic

1 solution

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