Lesson Miscellaneous problems on Vieta's theorem

Miscellaneous problems on Vieta's theorem

Problem 1

The given equation is the standard form quadratic equation

    5x^2 - 23x - 4 = 0.    (1)


Therefore, according to Vieta's theorem,  

    a + b = ,    (2)

    ab = .       (3)



Next,  

     +  = .    (4)


The numerator in (4) is

    a^2 + b^2 = (a^2 + 2ab + b^2) - 2ab = (a+b)^2 - 2ab = 

                replace here a+b by   and replace ab by    based on (2),(3) and continue

              =  -  =  +  =  = .


Therefore

     +  =  =  =  = 35 = 35.5625.    ANSWER

Problem 2

The given equation is the standard form quadratic equation

    6x^2 - 16x + 7 = 0.    (1)


According to Vieta's theorem, if "a" and "b" are the roots of this quadratic equation, then

    a + b =  = ,     (2)

      ab  = .           (3)


Now,   +  = .


The numerator is

    a^3 + b^3 = (a+b)*(a^2 - ab + b^2) = (a+b)*((a^2+2ab+b^2)-3ab) = (a+b)*((a+b)^2-3ab) = (a+b)^3 - 3(a+b)*(ab).


Now replace here  (a+b)  by   8/3  based on  (2)  and replace ab by  7/6  based on (3).

You will get  


    a^3 + b^3 =  -  =  -  =  -  =  = .


Therefore,   =  =  =  = .


ANSWER.   = .

Problem 3

According to Vieta's theorem, we have

    a + b = 5,

    ab = 3.


Next,   = .


In the numerator,  

    a^3 + b^3 = (a+b)*(a^2-ab+b^2) = (a+b)*((a^2 +2ab+b^2)-3ab) = (a+b)*((a+b^2-3ab) = (a+b)^3 - 3ab*(a+b) =

                substitute here a+b = 5, ab = 3 and continue

              = 5^3 - 3*3*5 = 125-45 = 80.


Therefore,  

     +  =  = .


Also,  . =  = ab = 3.


So, due to Vieta's theorem (again), the desired quadratic equation has 
the coefficient    at x  and  the constant term 3.


ANSWER.  The desired quadratic equation is  x^2 -  + 3 = 0.

Problem 4

Let p and q be the roots of the polynomial x^2 - x - 3.

Due to Vieta's theorem, 

    p + q = 1,        (1)

    pq = -3.          (2)


According to the problem, p and q are also the roots of the polynomial x^3 - bx^2 - c. 
Let r be the third root of this polynomial.

Then, due to Vieta's theorem for polynomial x^3 - bx^2 - c

    p + q + r = b,          (3)

    p*q + p*r + q*r = 0,    (4)   (coefficient at x in polynomial x^3 - bx^2 - c)

    p*q*r = c.              (5)


In (3), replace p+q by 1, based on (3).  In (5), replace p*q by -3, based on (2).
Then from (3) an (4) you will have

    1 + r = b,              (6)

    -3r =  c,               (7) 


In equation (4), replace p*q  by -3, based on (2).  Then equation (4) takes the form

    -3 + pr + qr = 0,  

or

    p*r + q*r  = 3,

    (p + q)*r  = 3.


In the last equation, replace (p + q) by 1,  based on (1).  You will get

    1*r = 3,  i.e.  r = 3.


Now from  (6)  b = 1 + r = 1 + 3 = 4;

    from  (7)  c = -3r = -3*3 = -9.


ANSWER.  In polynomial  x^3 - bx^2 - c, coefficients  are b = 4,  c = -9.

Problem 5

Notice that 

    p^2*qrs + pq^2*rs + pqr^2*s + pqrs^2 = pqrs*(p + q + r + s).


Also notice that the leading coefficient of the polynomial at x^4 is 1.


Due to Vieta's theorem

    pqrs          = -31  (the product of the roots is equal to the constant term)

    p + q + r + s =  -2  (the sum of the roots is the coefficient at x^3 with the opposite sign).


Therefore,

    p^2*qrs + pq^2*rs + pqr^2*s + pqrs^2 = (-31)*(-2) = 62.


ANSWER.  p^2*qrs + pq^2*rs + pqr^2*s + pqrs^2 = 62.

Problem 6

Notice that 

    p^2 + q^2 + r^2 + s^2 = (p + q + r + s)^2 - 2*(pq + pr +ps + qr + qs + rs)   (standard decomposition for the square of a sum)


Also notice that the leading coefficient of the polynomial at x^4 is 1.


Due to Vieta's theorem

    pq + pr + ps + qr + qs + rs = 16  (the sum of in-pairs products of the roots is the coefficient at x^2)

    p + q + r + s =  -2               (the sum of the roots is the coefficient at x^3 with the opposite sign).


Therefore,

    p^2 + q^2 + r^2 + s^2 = (-2)^2 - 2*16 = 4 - 32 = -28.


ANSWER.  p^2 + q^2 + r^2 + s^2 = -28.   

         (by the way, since the sum of squares is negative, it means that some roots are complex numbers).

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