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Quadratic Equations: Solutions by Formulae

Formula for solving a quadratic equation
Sometimes factorization of ax2 + b x + c is difficult.

Then, how do we solve ax2 + b x + c = 0?

We can derive a formula for solving a x + b x + c = 0

where a ≠ 0, b,c ε R

ax2 + b x + c = 0

ax2 + b x = - c

a(x2 + b/a x) = - c

x2 + (b/a) x = - c /a       ______(1)

The left-hand side of (1) is x2 + (b/a) x = x2 + 2 *  x *  b/2a

To make it a perfect square, add ( b/2a )2 = b2/ 4a2 on either side of (1)

x2 + 2 *  x *  b/2a + ( b/2a )2 = b 2 /4a2 - c/a = ( b2 - 4ac ) /4a2
 
(x + b/2a)2 = ( b2 - 4ac )/ 4a2
 
x + b/(2a) = (± ) / (2a)
     
x = (-b/(2a) ± ) /(2a)
     
x = (-b ± ) /(2a)
 
Thus, the roots of ax2 + b x + c = 0 are
 
x = (-b + ) / (2a) or x = (-b - ) / (2a)
 
We observe that the equation ax2 + b x + c = 0 has two roots and
 
they are (-b ± ) / (2a).

Relation between roots and coefficients of a quadratic equation
If the roots of ax2 + b x + c = 0 are p, q then
 
p = (-b + ) / (2a) and q = (-b - ) / (2a)
 
The sum of the roots
 
 
         = - 2b/ 2a

         = - b/a

The sum of the roots p + q = -b/a = -(coefficient of x / coefficient of x2)

The product of the roots =
 
p *  q = *  
 

        = { + b2 - ( b2 - 4ac) } / 2a *  2a

        = 4ac / 4a2 = c/a = constant term / coefficient of x2

Nature of the roots of a quadratic equation
The roots of a x2 + b x + c = 0 are
 
p =
 
q =
, where a ≠ 0, b, c ε R
 

Δ= b2 - 4ac is called the discriminant of ax2+ b x + c = 0.
( Δ is read as “delta” )

The nature of the roots, namely a, b, depend on Δ.

Since b2 - 4ac = 0 is a real number, we have the following possibilities.

b2 - 4ac = 0 or b2 - 4ac > 0 or b2 - 4ac < 0.

In each of these cases, we observe the nature of the roots:

  1. If b2 - 4ac = 0, i.e., Δ = 0 then
    p = -b/2a q = -b/2a, i.e., the two roots are real and equal.
    Thus, ax2 + b x + c = 0 has real and equal roots if Δ = 0.

  2. If Δ > 0, then the roots are real and distinct.

  3. If Δ < 0, the square root Δ is not real but is an imaginary number. Therefore, a, b are imaginary when Δ is negative.

We will learn later that the set containing all real and imaginary numbers is called the set of complex numbers. Hence, roots are complex when Δ < 0.

Example 1
Solve x2 - 6x + 5 = 0 using the formula.

Solution:
Comparing the given equation with ax2 + bx + c = 0, we get a = 1, b = - 6, c = 5
 

Therefore, roots are 		 

     
 

=

     
 

=

     
 

=

( 6 ± 4 )/2 = 5 or 1

Example 2
Find the sum and product of the roots of

9x2 + 4x - 11 = 0.

Solution:

Suppose a, b are the roots of the given equation. Then

a + b = -Coefficient of x / Coefficient of x2 = -4/9

a *  b = Constant term / Coefficient of x2 = -11/9

Example 3
If a, b are the roots of x2 - px + q = 0 then find the value of

a / b + b /a.

Solution:

a, b being roots of x2 - p x + q = 0

we have a + b = p, ab = q

Therefore, a / b + b / a = ( a2 + b2 ) / ab
                              = { (a + b)2 - 2ab } / ab = (p2 - 2q) /q

Example 4
If one root of x2 - ( p - 1 ) x + 10 = 0 is 5, then find the value of p and second root.

Solution:

Let the roots be 5, b

Then 5 + b = p - 1; 5b = 10

Therefore, b = 2

Therefore, 5 + 2 = p - 1 i.e., p = 8

Therefore, p = 8 and the second root is 2.

Example 5
Find the quadratic equation whose roots are 1/2, 3/2

Solution:

a + b = ½ + 3/2 = 2, a b = 1/2  3/2 = ¾

A quadratic equation with roots a b is x2 - ( a + b ) x + ab = 0

Therefore, the required equation is x2 - 2x + ¾ = 0 or 4x2 - 8x + 3 = 0

Example 6

Find the nature of the roots of the equations given below.

  1. x2 - 5x + 6 = 0

  2. x2+ 4x + 5 = 0

  3. x2+ 2x - 1 = 0

Solution:

  1. Comparing x2 - 5x + 6 = 0 with ax2+ b x + c = 0, we get a = 1, b = - 5, c = 6

    Δ = b2 - 4ac = 25 - 24 = 1
 

Therefore, the roots are

 
  1.            = ( 5 ± 1 ) / 2 = 3 or 2 real and distinct.

  2. In (ii), a = 1 b = 4 c = 5

    Δ= b2 - 4ac = 16 - 20 = -4 < 0
 

Therefore roots are a, b =

   

                                   =

   

                                   =

 
  1. In (iii) a = 1, b = 2, c = -1

    Δ = b2 - 4ac = 4 + 4 = 8 > 0
 

The roots are

   

            =

   

            =

-1 ± √2
 
which are real and irrational.
 

Try these questions

I. Using the formula for roots solve the following equations.

1.

x2- 4x - 12 = 0

2.

x2+ x - 42 = 0

3.

x2+ 16x + 48 =0

4.

3x2+ 2x - 8 = 0

5.

10x2- 7x - 12 = 0

6.

8 - 5x2 - 6x = 0 or - 5x2 - 6x + 8 = 0

7.

16x - 15 - 4x2 = 0

8.

6x2 - 13x - 63 =0

9.

12x2 + 3x - 99 = 0

10.

2x2 + 3x - 3 = 0

 

II. Find the sum and product of the roots of the equations given below
1.

3x2 + 2x + 1 = 0

2.

px2 - r x + q =0

3.

x2 - px + pq =0

4.

x2 +m x + m n = 0

5.

x2 - p x + q = 0
 

III.

If a, b are the roots of x2 - px + q = 0 the find the values of

( i ) a3 + b3

( ii ) a2/b + b2/a

( iii ) 1/a3 + 1/b3
 

I. Answers to Practice Problems
1.

x2 - 4x - 12 = 0


x =


Comparing the given equation with ax2 + bx + c = 0

We have a = 1; b = - 4 and c = - 12

b2 - 4ac = ( - 4 )2 - 4 1 ( - 12)

             = 16 + 48 = 64


Therefore x = { - ( - 4)

 } / 2 1


                  = 4 ± 8/2

x = 4 + 8 /2 or 4 - 8 /2 Therefore x = 6 or - 2
 

2.

x2 + x - 42 = 0


x =


Comparing the given equation with ax2+ b x + c = 0

We have a = 1, b = 1, c = - 42

b 2- 4ac = (1)2 - 4.1 ( - 42)

             = 1 + 168 = 169 > 0


Therefore x = { - 1 ±

 } / 2 1


x = - 1 ± 13/2

x = - 1 + 13/2 or - 1 - 13/2

x = 12/2 or - 14/2

Therefore x = 6 or - 7
 

3.

x2+ 16x + 48 =0


x =


Comparing the given equation with ax2 + b x + c = 0

We have a = 1, b = 16, c = 48

b2 - 4ac = (16)2 - 4  1* 48

             = 256 - 192 = 64

Therefore x = { - 16 ± √64} / 2 *1


x = - 16 + 8/2

x = - 16 + 8/2 or - 16 - 8/2

x = - 8 /2 or - 24/2

x = - 4 or - 12.
 

4.

3x2 + 2x - 8 = 0


x =


Comparing the given equation with ax2+ b x + c = 0

We have a = 3, b = 2, c = - 8

b2 - 4ac = (2)2 - 4 *3 * ( - 8)

              = 4 + 96 = 100

Therefore x =


x = - 2 ± 10/6

x = - 2 + 10/6 or - 2 - 10/6

x = 8/6 or - 12/6

x = 4/3 or - 2.
 

5.

10x2 - 7x - 12 = 0


x =


Comparing the given equation with ax2 + b x + c = 0

We have a = 10; b = - 7; c = - 12

b2 - 4ac = ( - 7)2 - 4  10  ( - 12)

              = 49 + 480 = 529

Therefore x =


x = 7 ± 23/20

x = 7 + 23/20 or 7 - 23/20

x = 30/20 or -16/20

Therefore x = 3/2 or - 4/5.
 

6.

8 - 5x2 - 6x = 0 or - 5x2- 6x + 8 = 0

Multiplying both sides by -

We have 5x2 + 6x - 8 = 0

Comparing the given equation with ax2 + b x + c = 0

We have a = 5; b = 6; c = - 8

b2 - 4ac = 62 - 4 *5 * ( - 8)

             = 36 + 160 = 196


x =



Therefore x =


x = - 6 ± 14/10

x = - 6 + 14/10 or - 6 - 14/10

x = 8/10 or -20/10

x = 4/5 or - 2.
 

7.

16x - 15 - 4x2 = 0

Multiplying both sides by -

4x2 - 16x + 15 =0

Comparing the given equation with ax2 + b x + c = 0

We have a = 4, b = - 16, c = 15

b2 - 4ac = ( - 16 )2 - 4 * 4 * 15

= 256 - 240 = 16


x =



Therefore x =


x = 16 + 4 / 8

x = 16 + 4/ 8 or 16 - 4/8

x = 20/8 or 12/8

x = 5/4 or 3/2
 

8.

6x2 - 13x - 63 =0

Comparing the given equation with ax2+ b x + c = 0

We have a = 6, b= - 13; c = - 63

b2 - 4ac = ( - 13)2 - 4 * 6 * ( - 63)

              = 169 + 1512

              = 1681


x =




Therefore x =


x = 13 ± 41/12

x = 13 + 41/12 or 13 - 41/12

x = 54 /12 or - 28/12

x = 9/2 or - 7/3
 

9.

12x2 + 3x - 99 = 0

Comparing the given equation with ax2+ b x + c = 0

a = 12, b = 3; c = -99

b2 - 4ac = (3)2 - 4  12  ( - 99)

             = 9 + 4752 = 4761


x =




Therefore x =


x = - 3 ± 69/ 24

x = -3 + 69/24 or - 3 - 69/24

Therefore = 66/ 24 or -72/24

x = 11/4 or - 3
 

10.

2x2 + 3x - 3 = 0

Comparing the given equation with ax2+ b x + c = 0

a = 2, b = 3, c = - 3

b2 - 4ac = (3)2 - 4 *2 * ( - 3)

             = 9 + 24 = 33


x =




Therefore x =


x = (-3 ± √33) / 4
 

II. Find the sum and product of the roots of the equations given below.
1.

3x2 + 2x + 1 = 0

Comparing the given equation with ax2+ b x + c= 0

We have a = 3, b = 2, c = 1,

Sum of roots = - b/a = - 2/3

Product of the roots = c/a = 1/3.
 

2.

px2 - r x + q =0

Comparing the given equation with ax2 + b x + c =0

We have a = p, b = - r, c = q

Some of roots = - b/a = - ( - r)/p = r/p

Product of the roots = c/a = q/p
 

3.

x2 - px + pq = 0

Comparing the given equation with ax2 + b x + c = 0

We have a = 1, b = - p, c = pq

Sum of root = - b/a = - ( - p) /1 =p

Product of the roots = c / a = pq/1 = pq
 

4.

x2 + m x + m n = 0

Comparing the given equation with ax2 + b x + c = 0

We have a = 1,b= m, c = mn

Sum of roots = - b/a

                      = - m/1 = - m

Product of the roots = c/a = mn /1 = mn.
 

5.

x2 - p x + q = 0

Comparing the given equation with ax2 + b x + c= 0

We have a = 1, b = - p, c = q

Sum of roots = - b/a = - (- p)/1 =p

Product of the roots =c/ a = q /1= q.
 

III.

The given equation is x2 - p x + q = 0

Let a, b be the roots of the given equation.

Sum of roots = a + b = - b/a = - ( - p)/1 =p

Product of roots = ab = c/a = q/1 = q

i). a3+ b3 = (a + b)3 - 3a b (a + b)

Recall: (a + b)3= a3 + b3 + 3ab (a + b)]

Substituting for a + b and ab.

We have a3+ b3 = (p)3 - 3. q (p) = p3 - 3pq



ii) a2/b + b2 /a = a3 + b3/ab

From (i) above we have a3 + b3= p3 - 3pq also a b = q

Therefore a2 / b + b2/ a = a3 + b3/ab = p3 - 3pq/q




iii) 1/a3 + 1/b3 = b3+ a3/a3b3 = a3+ b3/(ab)3

From (i) above we have a3 + b3 = p - 3pq

Therefore 1/a3+ 1/b3= a3+ b3/(ab)3= p3- 3pq/(q)3 = p3 - 3pq /q3

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