Quadratic Equations – – Sum & Product of Roots SS2 Further Mathematics Lesson Note

FINDING QUADRATIC EQUATION GIVEN SUM AND PRODUCT OF ROOTS CONDITION FOR EQUAL ROOTS, REAL ROOTS AND NO ROOT

We recall that if ax² + bx + c = 0, where a, a and c are constants such that a ≠ 0, then,

x  = -b + ✓b² – 4ac           

                  2a

x = -b – ✓b² – 4ac

                  2a

Suppose we represent these distinct roots by α and β; thus:

α = -b + ✓b² – 4ac           

                  2a

and

β = -b – ✓b² – 4ac

                  2a

We may also put D = b2 – 4ac, so that

α= + (b² + ✓D)

                2a

β = – (b² – ✓D)

              2a

Sum of roots

α + β = (+b + ✓D) + (-b – ✓D)

                  2a                 2a

= -2

   2b

= -b

    a

Products of roots

αβ = (-b + ✓D) × (-b – ✓D)

               2a                2a

˸αβ = b² – D

4a²

= b2 – (b2 – 4ac)

  4a²

= 4ac

   4a²

= c/a

Hence, if ax² + bx + c = 0, where a, b and c are constants andα≠ 0 then α + β= ,

αβ = 

x2 + x– 42 = 0

then (x – 6) (x – 7) = 0

Hence the roots of the equation are 6 and -7. In general, if a quadratic equation factorizes into

(x – α) (x – β) = 0

then α and β must be the roots of that equation. 

The general quadratic equation ax² + bx + c = 0 can also be written as:

x2 +  bx + c

          a     a        ………. i

If the roots of the equation are α and β then the above equation can be written as:

(x –α) (x – β) = 0

x2 – (α – β) x + αβ = 0   ……..ii

By comparing coefficients in equations (1) and (2)

-(α + β) = b

                  a

: α + β = -b

                a

andαβ = c

              d

The above consideration gives rise to two problems:

(a) Given a quadratic equation, we can find the sum and product of the roots.

(b) Given the roots, we can formulate the corresponding quadratic equation.

The quadratic equation whose roots are α and β is

x2 – (α + β) x + α β = 0

Find the sum and product of the roots of each of the following quadratic equations:

(a) 2×2 + 3x – 1 = 0

(b) 3×2 – 5x – 2 = 0

(c) x2 – 4x – 3 = 0

(d) ½ x2 – 3x – 1 = 0

Solution

(a) 2×2 + 3x – 1 = 0

a = 2; b = 3; c = -1

Let α and β be the roots of the equation, then

α + β= -b = -3

              a      2

α β = c  = -1

          a      2

(b) 3x² – 5x – 2 = 0

a = 3; b = -5; c = -2

Let α and β be the root of the equation, then

α + β = -b = 4

              c      1

α β = c   = -3

          a

(c) x2 – 4x – 3 = 0

a = 1; b = 4; c = -3

Let α and β be the root of the equation, then

α + β = 

α β = 

(d) ½ x2 – 3x – 1 = 0

a = ½, b = -3, c = -1

Let α and β be the root of the equation, then

α + β = 

α β =  = -2

Find the quadratic equation whose roots are:

(a) 3 and -2

(b) ½ and 5

(c) -1 and 8

(d)¾ and ½ 

Solution

The quadratic equation whose roots are α and β is x2 – (α + β) x +α β = 0.

(a) α + β = 3 – 2 = 1, α β = 3 (-2) = -6

: The quadratic equation whose roots are 3 and -2 is x2 – x – 6 = 0.

(b) α β =   α β = 

:The quadratic equation whose roots are 

x2– 

or 2×2 – 11x + 5 = 0

(c) α+ β = 7,  α β = -8

:α β = 7,α β = -8

:The quadratic equation whose roots are -1 and 8 is x2 – 7x – 8 = 0.

(b) α+ β =  α β = 

:The quadratic equation whose roots are ¾ and ½ is

x2– 

or 8×2 – 10x + 3 = 0