Lesson Problems on quadratic equations to solve them using discriminant

Problems on quadratic equations to solve them using discriminant

Problem 1

Quadratic equation  ax^2 + bx + c = 0  does not have real solutions if and only if
its discriminant d = b^2 - 4ac  is negative.


In our case, the discriminant is

    d = b^2 - 4ac = (-1)^2 - 4*2*3 = 1 - 24 = -23.


The discriminant is negative- hence, this equation  2x^2 - x + 3 = 0  does not have real solutions.

Problem 2

The starting equation is

    13x^2 - kx - 9 = 0.    (1)


Quadratic equation (1) has at least one real solution if and only if its discriminant is non-negative.

Write the discriminant

    d = b^2 - 4ac = (-k)^2 - 4*13*9 = k^2 - 468.    (2)


Discriminant (2) is non-negative if

    k^2 - 468 >= 0,  or  k^2 >= 468,  |k| >= ,  |k| >= ,  |k| >= .    (3)


Since we are looking for positive values of 'k',  inequality (3)  takes the form  k >= .


ANSWER.  The smallest positive value of 'k' is   = 21.6333  (approximately).

Problem 3

This problem can be equivalently re-phrased in this way


    Let 'c' be a real number.  Find the maximum value of 'c' such that 
    equation  2x^2 = x+c  has at most one solution in real numbers.


It is equivalently to the condition that the discriminant
of equation  2x^2 - x - c = 0  is non-positive real number.


The discriminant of the last equation is 

    d = b^2 - 4ac = (-1)^2 - 4*2*(-c) = 1 + 8c.


Inequality  d <= 0  for the discriminant is

    1 + 8c <= 0.


It implies

    c <= -1/8.


The maximum value of 'c', satisfying this inequality, is  -1/8.


ANSWER.  The maximum value of 'c', satisfying the imposed condition is  -1/8.

Problem 4

The given quadratic function 3-4k - (k+3)x - x^2  has the leading coefficient -1 at x^2,
so it represents a parabola opened downward.


In order for this quadratic polynomial be negative at all real values of x, the necessary and 
sufficient condition is that the discriminant

    d = b^2 - 4ac 

be negative.  Then the square function has no real roots and remains negative for all real values of x.


So, we write the discriminant

    d = (-(k+3))^2 - 4*(-1)*(3-4k) = (k+3)^2 + 4*(3-4k) = k^2 + 6k + 9 + 12 - 16k = k^2 - 10k + 21.


It can be factored

    d = (k-7)*(k-3).


So, the discriminant is negative

    (k-7)*(k-3) < 0


if and only if  the parameter "k" is between the roots  3 < k < 7,
when the factor (k-7) is negative, while the factor (k-3) is positive.


At this point, the problem is fully solved.


ANSWER.  The range of values of k for which the expression  3 - 4k - (k+3)x - x^2  will be negative 

         for all real values of x  is  3 < k < 7.

Problem 5

The given inequality is 

    4x^2 + nx + 121 < 0.


It has no real solutions if and only if the discriminant 
of the quadratic polynomial in the left side is negative

    d = b^2 - 4ac < 0

    n^2 - 4*4*121 < 0

    n^2 < 16*121

    |n| <  

    |n| < 4*11

    |n| < 44

    -44 < n < 44.


This inequality has  43 + 1 + 43 = 87 solutions in integer numbers
(43 negative, one zero and 43 positive integer solutions).


ANSWER.  There are 87 different possible integer values of n such that
         the given inequality has no real solutions.

Problem 6

Left side  x^3 + y^3  can be composed  as  (x+y)*(x^2 - xy + y^2).

The factor  (x+y)  can be replaced by 8,  based on the first equation.


After that, second equation can be written in this form

    8(x^2 - xy + y^2) = 200 - 4(x^2 + y^2),

or

    12(x^2 + y^2) - 8xy = 200,

    12(x+y)^2 - 24xy - 8xy = 200,

    12(x+y)^2 - 32xy = 200.


Again, we can replace here  (x+y)  by 8, and we get then

    12*8^2 - 32xy = 200,

    32xy = 12*64 - 200 = 568,

      xy = 568/32 = 17.75.


But under the condition x+y = 8,  well known minimax property says 
that the product xy may have the maximum value of  4*4 = 16  at  x = y = 8/2 = 4.


In other words, under given conditions, the original system has no real solutions.


So, the given system of equations does not have solutions in real numbers..