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Nature of Roots

(1) Equation ax2 + bx + c = 0; a ¹0 will have real roots only when D = b2 - 4ac > 0.
(2) When D = b2 - 4ac<0, then the equation ax2 + bx + c = 0 will have no real roots.
(3) When D = b2 - 4ac = 0, then the roots will be real and equal, each being equal to .
 

Sum and Product of the Roots of a Quadratic Equation
Let there be a quadratic equation ax2+bx+c=0, whose discriminant, D=b2-4ac 0.
Then its two real roots
α and β will be written as
         
α= and β
Therefore, the sum of the roots of the equation,

α + β
        = 
and the product of the roots will be written as, 


α β
     =       
Therefore, the sum of the roots = and product of the roots = .

 

Example

Find the discriminant of the following equations:
(i) 2x2 + 5x - 6 = 0            (ii) 7x2 - 2x + 6 = 0
(iii)             (iv) 3x2 - 5x + 2 = 0
(v) 6x2 - 7cx + 8 = 0  

Solution

(i) The given equation is 2x2 + 5x - 6 = 0
Here a = 2, b = 5 and c = -6
Therefore, discriminant D = b2 - 4ac
 =52 - 4(2)(-6) = 73
(ii) The given equation is
7x2-2x+6=0 in which
a=7, b=-2, c=6
D=b2-4ac=(-2)2-(4)(7)(6)=-164
(iii) 
a=
D = b2-4ac =  
Substitute
3 = 1.732                   
   =192+76.2 = 268.2
(iv) 3x2 - 5x + 2 = 0
Here a = 3, b = -5, c = 2
D = b2 - 4ac = (-5)2 - (4)(3)(2) = 1
(v) 6x2 - 7cx + 8 = 0
Here a = 6, b = -7c, c = 8
D = b2 - 4ac = (-7c)2 - (4)(6)(8)
   = 49c2 - 192.


 

Example

Comment on the nature of the roots of the following equations:
(i) 3x2 + 5x + 1 = 0              (ii) 4x2 - 2x + 3 = 0
(iii) 7x2 - 13x + 5 = 0            (iv) 
(v) x2 + 6x + 9 = 0

Solution

(i) The given equation is
         3x2 + 5x + 1 = 0
Here   a = 3, b = 5, c = 1
Discriminant D = b2 - 4ac
                     = (5)2 - (4)(3)(1) = 13
Here, D = 13>0. Hence the roots will be real and unequal.
(ii)  For the equation       4x2 - 2x + 3 = 0
       a = 4, b = -2, c = 3
       D = b2 - 4ac = (-2)2 - (4)(4)(3) = -44
Here D = -44 < 0
Therefore there are no real roots.
(iii) 7x2 - 13x + 5 = 0
      a = 7, b = -13, c = 5
      D = b2 - 4ac = (13)2 - (4)(7)(5) = 29
Here D>0. Therefore unequal and real roots will exist.
(iv) 
      a =
      D = b2 - 4ac = (-
         =7 + 36 = 43>0
... Unequal and real roots exists.
(v)  x2 + 6x + 9 = 0
      a = 1,b = 6,c = 9
      D = b2 - 4ac = (6)2 -(4)(1)(9) = 0
      Therefore real and equal roots.


 

Example

Find the value of p, for which the following equations have two equal roots.
(i)
(ii) 4x2 - 8x + p = 0

Solution

(i) The given equation is
Here,


For real and equal roots, D = 0
            
      where
p is any real number.
(ii)     4x2 - 8x + p = 0
         a = 4, b = -8, c = p
         D = b2 - 4ac, i.e. D = (-8)2 - 4.4. p
            =64 - 16p
For real and equal roots, D = 0

         64 - 16p = 0 or p = 4.
where p is any real number.


 

Example

In the following, determine whether the given quadratic equations have real roots and if so, find the roots.
(i) x2 + 7x + 10 = 0                     (ii) 4x2 - 3x - 10 = 0
(iii) 8x2 + 8x + 1 = 0                   (iv) 2x2 + 4x + 1 = 0
(v) 

Solution

(i) The given equation is x2 + 7x + 10 = 0
Here, a = 1, b = 7, c = 10
D = b2 - 4ac=72 -(4)(1)(10) = 9 >0
As D>0, therefore real roots are possible.
Again, x2 + 7x + 10 = 0

         x2 + 5x + 2x +10 = 0

       x(x + 5) + 2(x + 5) = 0
              (x + 5)(x + 2) = 0
Therefore, x = -5 and -2.
(ii) The given equation is
     4x2 - 3x - 10 = 0
and a = 4, b = -3, c = -10
      D = b2 -4ac =(-3)2 - (4)(4)(-10)
         = 9 + 160 =169
As D>0, therefore real roots will exist. The two real roots of the equation will be written as,


α =      β =

α =           β

α = 2                    β =
Therefore, the roots are 2 and
(iii) 8x2 + 8x + 1=0
      a = 8, b = 8, c = 1
      D = b2 - 4ac = (8)2 - (4)(8)(1) = 32
As D>0, therefore the equation will have real roots given by,


α =

β =  

(iv) 2x2 + 4x + 1 = 0
     a = 2, b = 4, c = 1
     D = b2 - 4ac = (4)2 - (4)(2)(1) = 8.
As D > 0, so the equation will have real roots given by,


α

β

(v) 

     a = ,  b = -3, c = +
     D = (-3)2 - 4. . = 9 - 20 = -11
     As D < 0, no real roots are possible.

 
 

Example

Prove that, both the roots of the equation (x - p) (x - q) + (x - q) (x - r)+ (x - r) (x - p) = 0 are real, but they are equal only when p = q = r.

Solution

The given equation is
(x - p) (x - q) + (x - q) (x - r) + ( x - r) (x - p) = 0
3x2 - 2x(p + q + r) + (pq + qr + rp)=0
D = b2 - 4ac
   = [-2(p + q + r)]2 - [4 × 3 (pq + qr + rp)]
   = 4 (p + q + r)2 - 12(pq + qr + rp)
   = 4 (p2 + q2 + r2 - pq - qr - rp)
   = 2 (2p2 + 2q2 + 2r2 - 2pq - 2qr - 2rp)

   = 2(p2 - 2pq + q2 + q2 - 2qr + r2 + r2 - 2pr + p2)
   = 2 [(p - q)2 + (q - r)2 + (r - p)2]
0
Therefore, the roots of the equation are real. For equal roots of a quadratic equation D must be equal to zero.
Hence, 2[(p - q)2 + (q - r)2 + (r - p)2] = 0
This is possible only when,
       p - q = 0 or   p = q
and q - r = 0   or   q = r
and r - p = 0   or    r = p
i.e.  p = q = r
Therefore, the roots of the given equation are equal, only when p = q = r.

 

 




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