High School How to Find the Value of \( a \) for Evenly Spaced Roots in an Equation?

[anemone]

Determine the value for $a$ so that the equation $(x^2-1)(x^2-4)=a$ has four evenly spaced nonzero real roots.

--------------------
Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solutions:

1. MarkFL
2. magneto
3. lfdahl

Here is MarkFL's solution:

Spoiler

Let:

$$f(x)=\left(x^2-1\right)\left(x^2-4\right)-a$$

Since this is an even function, its roots will be symmetric across the $y$-axis. Let the positive roots therefore be:

$$x\in\{k,3k\}$$ where $0<k$

There results:

$$f(k)=\left(k^2-1\right)\left(k^2-4\right)-a=0$$

$$f(3k)=\left(9k^2-1\right)\left(9k^2-4\right)-a=0$$

This implies:

$$a=\left(k^2-1\right)\left(k^2-4\right)=\left(9k^2-1\right)\left(9k^2-4\right)$$

Solve for $k$:

$$k^4-5k^2+4=81k^4-45k^2+4$$

$$k^4-5k^2=81k^4-45k^2$$

Since $0<k$, we may divide through by $k^2$:

$$k^2-5=81k^2-45$$

$$80k^2-40=0$$

$$2k^2-1=0$$

$$k^2=\frac{1}{2}$$

Hence:

$$a=\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-4\right)=\frac{7}{4}$$

Thus, the function:

$$f(x)=\left(x^2-1\right)\left(x^2-4\right)-\frac{7}{4}$$

has 4 evenly spaced roots, given by:

$$x\in\left\{-\frac{3}{\sqrt{2}},-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},\frac{3}{\sqrt{2}}\right\}$$

Here is lfdahl's solution:

Spoiler

\[(x^2-1)(x^2-4)= a \Rightarrow (x^2)^2-5x^2 +4-a = 0\\\\ x^2= \frac{1}{2}(5\pm \sqrt{25-4(4-a)})\Rightarrow x^2 = \frac{5}{2}\pm \sqrt{\frac{9}{4}+a} \;\;\;\; -\frac{9}{4} \le a \le 4\]
So $x \in \left \{ \pm x_1,\pm x_2 \right \} = \left \{ \pm \sqrt{\frac{5}{2}+ \sqrt{\frac{9}{4}+a}} ,\;\; \pm \sqrt{\frac{5}{2}- \sqrt{\frac{9}{4}+a}}\right\}$
Evenly spaced roots means:
\[x_1-x_2 = 2x_2 \Rightarrow x_1 = 3x_2\]
This determines $a$:
\[x_1 = 3x_2 \Rightarrow \sqrt{\frac{5}{2}+ \sqrt{\frac{9}{4}+a}}= 3\sqrt{\frac{5}{2}- \sqrt{\frac{9}{4}+a}}\]
I get: $a=\frac{7}{4}$