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Three roots of the equation $x^4-px^3+qx^2-rx+s=0$ are $\tan A$, $\tan B$ & $tan C$ where $A$, $B$, $C$ are the angles of a triangle.The fourth root of the biquadratic is

  1. $(p-r)$/$(1-q+s)$
  2. $(p-r)$/$(1+q-s)$
  3. $(p+r)$/$(1-q+s)$
  4. $(p+r)$/$(1+q-s)$

I tried using theory of equations and the identity that in a triangle $\tan A+\tan B+\tan C = \tan A \tan B\tan C$ but could not find the fourth root.This is my solution

My answer doesn't match any option please tell me what have I done wrong and what is the correct way to do this

miracle173
  • 11,049
Gem
  • 792

3 Answers3

2

HINT:

Let $t$ be fourth root.

Using Vieta's formulas

$$q-s$$

$$=\tan A\tan B+\tan B\tan C+\tan C\tan A+t(\tan A+\tan B+\tan C)-t(\tan A\tan B\tan C)$$

$$=\tan A\tan B+\tan B\tan C+\tan C\tan A$$

Again,

$$p-r$$

$$=\tan A+\tan B+\tan C+t-\{\tan A\tan B\tan C+t(\tan A\tan B+\tan B\tan C+\tan C\tan A)\}=t(1-\underbrace{\tan A\tan B+\tan B\tan C+\tan C\tan A})$$

lab bhattacharjee
  • 274,582
1

Assume fourth root is $\tan D$

$\tan(A+B+C+D)=(S_1 - S_3)/(1-S_2+S_4)$

$\tan (A+B+C+D) = \tan (π+D) = \tan (D)$

So we get the fourth root as $(S_1 - S_3)/(1-S_2+S_4)$ which simplifies to $(p-r)/(1-q+s)$

$S_n$ is sum of tangent of angles taken $n$ at a time

miracle173
  • 11,049
Gem
  • 792
0

For the lazy ones I have this piece of code in Maxima. Here is an online version of Maxima. Note that maxima uses : to assigne values to variables.

display2d:false;
t:2.5;
A:%pi/3.0;
B:%pi/4.0;
C:%pi-A-B;
e:(x-tan(A))*(x-tan(B))*(x-tan(C))*(x-t),expand,numer;
p:-coeff(e,x,3);
q:coeff(e,x,2);
r:-coeff(e,x,1);
s:coeff(e,x,0);
(p-r)/(1-q+s);
(p-r)/(1+q-s);
(p+r)/(1-q+s);
(p+r)/(1+q-s);

So we use $A=\pi/3, B=\pi/4, C=\pi-A-B$ and the 4th root $t=2.5$ which produces the following output

(%i1) display2d:false;

(%o1) false
(%i2) 
t:2.5;

(%o2) 2.5
(%i3) 
A:%pi/3.0;

(%o3) 0.33333333333333*%pi
(%i4) 
B:%pi/4.0;

(%o4) 0.25*%pi
(%i5) 
C:%pi-A-B;

(%o5) 0.41666666666667*%pi
(%i6) 
e:(x-tan(A))*(x-tan(B))*(x-tan(C))*(x-t),expand,numer;

(%o6) x^4-8.964101615137757*x^3+28.08845726811991*x^2-36.28460969082654*x
         +16.16025403784439
(%i7) 
p:-coeff(e,x,3);

(%o7) 8.964101615137757
(%i8) 
q:coeff(e,x,2);

(%o8) 28.08845726811991
(%i9) 
r:-coeff(e,x,1);

(%o9) 36.28460969082654
(%i10) 
s:coeff(e,x,0);

(%o10) 16.16025403784439
(%i11) 
(p-r)/(1-q+s);

(%o11) 2.499999999999999
(%i12) 
(p-r)/(1+q-s);

(%o12) -2.113248654051871
(%i13) 
(p+r)/(1-q+s);

(%o13) -4.140544456622766
(%i14) 
(p+r)/(1+q-s);

(%o14) 3.499999999999999

Comment:

  • %i is an input line %o is an output line
  • %i1 can be ignored, it isabout the format of the output
  • %i2 - %i5: values are assigned to the variables t,A,B,C. The values of at,A,B are arbitrary, C is the complement of A+B to $\pi$.
  • %i6: e is the polynomial constructed according to Vieta, expand and numer are flags that adises Maxima to actually do the multiplication and use numeric values for $\pi$.
  • %i7 - %i10 assignes the coeffizient of the polynomial expression e to the variables p,q,r,s
  • %i11 - %i14 calculates the expressions given by the multiple choice test

If we check the output %o11, %o12, %o13, %o14 we see that $(p-r)/(1-q+s)$ produces approximately the requested value $t$.

miracle173
  • 11,049