Lesson Using Vieta's theorem to solve quadratic equations and related problems

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Using Vieta's theorem to solve quadratic equations and related problems

Problem 1

If "a" and "b" are the roots of the quadratic equation x%5E2+-+3x+-1

= 0, find a%5E2+%2B+b%5E2

.

Solution 1

According to Viets's theorem,   a + b = 3   and   ab = -1.


Hence,   =  =  = 9 + 2 = 11.


Answer.    = 11.


Notice that we do not need to solve the given quadratic equation explicitly to get the answer.

Problem 2

If two roots of the quadratic polynomial 7x%5E2%2B3x%2Bk

are

, find k.

Solution

 is the product of the roots (Vieta's theorem):

 =  =  = .


Hence, k =  = 11.

Problem 3

The roots of the quadratic equation z%5E2+%2B+az+%2B+b

= 0 are 2 - 3i and 2 + 3i. What is a+b?

Solution

-a = (2-3i) + (2+3i) = 4  ====>  a = -4.

b = (2-3i)*(2+3i) =  = 4 + 9 = 13.

====>  a + b = -4 + 13 = 9.


The Vieta's theorem.


Answer. a + b = 9.

Problem 4

Suppose "a" and "x" satisfy x%5E2+%2B+%28a-%281%2Fa%29%29x+-+1

= 0. Solve for "x" in terms of "a".

Solution

Solution 1 (the Vieta's theorem)

The Vieta's theorem says:  if  p and q are the roots of a quadratic equation   = 0  then  u = -(p+q)  and  v = pq.

The opposite is also TRUE:  if  u = -(p+q)  and  v = pq  then p and q  are the roots of the quadratic equation     = 0.


Now look into your equation and notice that  the numbers  -a  and    give   when summed up and -1 when multiplied.


Hence,  the numbers "-a" and   are the roots of your equation.

Solution 2 (Factoring)

Factor your polynomial:    = .


Hence, the roots are "-a" and .

Problem 5

Without solving the given equation, find an equation whose roots are the squares of the roots of x%5E2+%2B+4x+%2B+2

= 0.

Solution

Let "s" and "t" be the roots of the quadratic polynomial

p(x) =  

with the leading coefficient 1 at . Then the Vieta's theorem says:

s + t = -p   and  s*t = q.


For the given equation 

 = 

it means that if "s" and "t" are its roots, then

s + t = -4,    (1)    and
s*t    = 2.    (2)

Next, if (1) and (2) are held, then

 =  =  = 16;  hence,  = 16 - 2s*t = 16 - 2*2 = 12,   and

 =  =  = 4.


Hence, by applying the Vieta;s theorem once again, you see that  and  are the roots of the polynomial

g(x) = .

It is the answer to the problem's question.   The problem is solved !!


Notice,  we get the answer without solving the original equation.
Exactly as assigned by the condition.
All we did was manipulating with coefficients and using the Vieta's theorem.

Problem 6

Determine all values of "c" so that (2x+1) is a factor of 2x%5E2%2B7x%2Bc

.

Solution

1.  The fact that 2x+1 is the factor of the polynomial p(x) = 2x^2+7x+c  is equivalent to divisibility 
    the polynomial p(x) by .


2.  Divisibility of the polynomial p(x) by  is equivalent to the fact that  is the root of the polynomial p(x):

     = 0.     (the "Remainder theorem")


3.  The fact that  is the root of the polynomial p(x) means that

     = 0,   or    = 0,   or     c =  =  =  = 3.


Answer.  c = 3.

Problem 7

is a root of the equation 2x^2-2kx+4 = 0, find the value of "k".

Solution

The given equation

    2x^2 - 2kx + 4 = 0 

is equivalent to  (after dividing both sides by 2)


    x^2 - kx + 2 = 0.      (1)


We are given that    is the root of the original equation; hence, the equation (1) with the leading coefficient 1 has this root, too.


Then, applying the Vieta's theorem, the other root of the equation (1) is


     = . =  =  = .


Thus we know BOTH ROOTS of the equation (1) (even without solving it explicitly (!) ). They are


      and  .


Again, according to Vieta's theorem, the sum of these roots is the coefficient at x in equation (1) taken with the opposite sign:


    k =  +  = .


Answer.  k = .

Problem 8

If the equations x%5E2+-+ax+%2B+b

= 0 and

= 0 have one root in common and if the second equation has equal roots, then prove that ae = 2(b + f).

Solution

1.  If  = 0  has equal roots, it means that its discriminant d = e^2 - 4f  is zero:

     = 0,   or    = ,   or   = 2f.    (1)



2.  Then the merged root of the equation   = 0  is  

     =     ( <<<---=== it follows from the quadratic formula, for example )



3.  It implies that the common root of the two given equations is  ,  since the second equation has no other roots.



4.  Thus we know that  is one of the roots of the equation   = 0.

    Then, according to the Vieta's theorem, the other root of this equation is 

    ,  and the product of the roots is the constant term


     = b.



5.  The last equality is equivalent to 

     -  = 2b  <----->  ae =  +   <----->  ae = 2b + 2f  <----->  ae = 2(b+f).

    The transformation step before the last one uses the equality (1).


It is what has to be proved.

Problem 9

One solution of z^3+ A*z^2 - z - 14 = 0 is z = -2-root(3)*i, where A is a real number.
Find A and other two solutions.

Solution

Since the given equation is with real coefficients (given), its complex roots are conjugated complex numbers.

Since one such root is   =   (given), the other root is   = .


Vieta's theorem says that the product of three roots of the given equation equals to the constant term 
with the opposite sign. So

     = 14.


The product of the two complex conjugated numbers    and    is 

     = 4 + 3 = 7.


Therefore, the third root of the given equation is   =  = 2.


Now the coefficient A equals to the sum of the three roots with the opposite sign, due to the same Vieta's theorem

    A =  =  = -(-2-2+2) = 2.


ANSWER.  A = 2.  The other two roots are    and  2.

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