Quadratic Equations: Solutions by Formulae
Formula for solving a quadratic equation
Sometimes factorization of ax^{2} + b x + c is difficult.
Then, how do we solve ax^{2} + b x + c = 0?
We can derive a formula for solving a x + b x + c = 0
where a ≠ 0, b,c ε R
ax^{2} + b x + c = 0
ax^{2} + b x =  c
a(x^{2} + b/a x) =  c
x^{2} + (b/a) x =  c /a ______(1)
The lefthand side of (1) is x^{2} + (b/a) x = x^{2} + 2 * x * b/2a
To make it a perfect square, add ( b/2a )^{2} = b^{2}/ 4a^{2} on either side of (1)
x^{2} + 2 * x * b/2a + ( b/2a )^{2} = b 2 /4a2  c/a = ( b^{2}  4ac ) /4a^{2}
(x + b/2a)^{2} = ( b^{2}  4ac )/ 4a^{2}
x + b/(2a) = (± ) / (2a)
x = (b/(2a) ± ) /(2a)
x = (b ± ) /(2a)
Thus, the roots of ax^{2} + b x + c = 0 are
x = (b + ) / (2a) or x = (b  ) / (2a)
We observe that the equation ax^{2} + 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 ax^{2} + 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 = *
= { + b^{2}  ( b^{2}  4ac) } / 2a * 2a
= 4ac / 4a^{2} = c/a = constant term / coefficient of x^{2}
Nature of the roots of a quadratic equation
The roots of a x^{2} + b x + c = 0 are
p =
q = , where a ≠ 0, b, c ε R
Δ= b^{2}  4ac is called the discriminant of ax^{2}+ b x + c = 0.
( Δ is read as "delta" )
The nature of the roots, namely a, b, depend on Δ.
Since b^{2}  4ac = 0 is a real number, we have the following possibilities.
b^{2}  4ac = 0 or b^{2}  4ac > 0 or b^{2}  4ac < 0.
In each of these cases, we observe the nature of the roots:

If b^{2}  4ac = 0, i.e., Δ = 0 then
p = b/2a q = b/2a, i.e., the two roots are real and equal.
Thus, ax^{2} + b x + c = 0 has real and equal roots if Δ = 0.

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

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 x^{2}  6x + 5 = 0 using the formula.
Solution:
Comparing the given equation with ax^{2} + 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
9x^{2} + 4x  11 = 0.
Solution:
Suppose a, b are the roots of the given equation. Then
a + b = Coefficient of x / Coefficient of x^{2} = 4/9
a * b = Constant term / Coefficient of x2 = 11/9
Example 3
If a, b are the roots of x^{2}  px + q = 0 then find the value of
a / b + b /a.
Solution:
a, b being roots of x^{2}  p x + q = 0
we have a + b = p, ab = q
Therefore, a / b + b / a = ( a^{2} + b^{2} ) / ab
= { (a + b)^{2}  2ab } / ab = (p^{2}  2q) /q
Example 4
If one root of x^{2}  ( 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 x^{2}  ( a + b ) x + ab = 0
Therefore, the required equation is x^{2}  2x + ¾ = 0 or 4x^{2}  8x + 3 = 0
Example 6
Find the nature of the roots of the equations given below.
x^{2}  5x + 6 = 0

x^{2}+ 4x + 5 = 0
x^{2}+ 2x  1 = 0
Solution:

Comparing x^{2}  5x + 6 = 0 with ax^{2}+ b x + c = 0, we get a = 1, b =  5, c = 6
Δ = b^{2}  4ac = 25  24 = 1
Therefore, the roots are
= ( 5 ± 1 ) / 2 = 3 or 2 real and distinct.
In (ii), a = 1 b = 4 c = 5
Δ= b^{2}  4ac = 16  20 = 4 < 0
Therefore roots are a, b =
=
=
In (iii) a = 1, b = 2, c = 1
Δ = b^{2}  4ac = 4 + 4 = 8 > 0
The roots are
=
=1 ± √2
which are real and irrational.
these questions
I. Using the formula for roots solve the following equations.
x^{2} 4x  12 = 0
x^{2}+ x  42 = 0
x^{2}+ 16x + 48 =0
3x^{2}+ 2x  8 = 0
10x^{2} 7x  12 = 0
8  5x^{2}  6x = 0 or  5x^{2}  6x + 8 = 0
16x  15  4x^{2} = 0
6x^{2}  13x  63 =0
12x^{2} + 3x  99 = 0
2x^{2} + 3x  3 = 0
II. Find the sum and product of the roots of the equations given below
3x^{2} + 2x + 1 = 0
px^{2}  r x + q =0
x^{2}  px + pq =0
x^{2} +m x + m n = 0
x^{2}  p x + q = 0
III. If a, b are the roots of x^{2}  px + q = 0 the find the values of
( i ) a^{3} + b^{3}
( ii ) a^{2}/b + b^{2}/a
( iii ) 1/a^{3} + 1/b^{3}
I. Answers to Practice Problems
x^{2}  4x  12 = 0
x =
Comparing the given equation with ax2 + bx + c = 0
We have a = 1; b =  4 and c =  12
b^{2}  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
x^{2} + x  42 = 0
x =
Comparing the given equation with ax^{2}+ 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

x^{2}+ 16x + 48 =0
x =
Comparing the given equation with ax^{2} + b x + c = 0
We have a = 1, b = 16, c = 48
b^{2}  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.

3x^{2} + 2x  8 = 0
x =
Comparing the given equation with ax^{2}+ b x + c = 0
We have a = 3, b = 2, c =  8
b^{2}  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.

10x^{2}  7x  12 = 0
x =
Comparing the given equation with ax^{2} + b x + c = 0
We have a = 10; b =  7; c =  12
b^{2}  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.
8  5x^{2}  6x = 0 or  5x^{2} 6x + 8 = 0
Multiplying both sides by 
We have 5x^{2} + 6x  8 = 0
Comparing the given equation with ax^{2} + b x + c = 0
We have a = 5; b = 6; c =  8
b^{2}  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.
16x  15  4x^{2} = 0
Multiplying both sides by 
4x^{2}  16x + 15 =0
Comparing the given equation with ax^{2} + b x + c = 0
We have a = 4, b =  16, c = 15
b^{2}  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
6x^{2}  13x  63 =0
Comparing the given equation with ax^{2}+ b x + c = 0
We have a = 6, b=  13; c =  63
b^{2}  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

12x^{2} + 3x  99 = 0
Comparing the given equation with ax^{2}+ b x + c = 0
a = 12, b = 3; c = 99
b^{2}  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

2x^{2} + 3x  3 = 0
Comparing the given equation with ax^{2}+ b x + c = 0
a = 2, b = 3, c =  3
b^{2}  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.
 3x^{2} + 2x + 1 = 0
Comparing the given equation with ax^{2}+ 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.
 px^{2}  r x + q =0
Comparing the given equation with ax^{2} + 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
 x^{2}  px + pq = 0
Comparing the given equation with ax^{2} + 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
 x^{2} + m x + m n = 0
Comparing the given equation with ax^{2} + 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.
 x^{2}  p x + q = 0
Comparing the given equation with ax^{2} + 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). a^{3}+ b^{3} = (a + b)^{3}  3a b (a + b)
Recall: (a + b)^{3}= a^{3} + b^{3} + 3ab (a + b)]
Substituting for a + b and ab.
We have a^{3}+ b^{3} = (p)^{3}  3. q (p) = p^{3}  3pq
ii) a^{2}/b + b^{2} /a = a^{3} + b^{3}/ab
From (i) above we have a^{3} + b^{3}= p^{3}  3pq also a b = q
Therefore a^{2} / b + b^{2}/ a = a^{3} + b^{3}/ab = p^{3}  3pq/q
iii) 1/a^{3} + 1/b^{3} = b^{3}+ a^{3}/a^{3}b^{3} = a^{3}+ b^{3}/(ab)^{3}
From (i) above we have a^{3} + b^{3} = p  3pq
Therefore 1/a^{3}3+ 1/b^{3}= a^{3}+ b^{3}/(ab)^{3}= p^{3} 3pq/(q)^{3} = p^{3}  3pq /q^{3}