SOLUTION TO QUADRATIC EQUATION

FURTHER MATHEMATICS SSS2 FIRST TERM

WEEK 1

SOLUTION TO QUADRATIC EQUATION

Contents

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+b2-4ac2a    or x = -b-b2-4ac2a

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

α = -b+b2-4ac2a

and

β-b-b2-4ac2a

We may also put D = b² – 4ac, so that

α= -b+D2a

β = -b-D2a

Sum of roots

α + β = (-b+D)2a + (-b-D)2a

= -22b

= -ba

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

αβ = ca,  we recall from 5.3 that by the method of factorization if

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 + bxa + ca =  0                                                …(1)

 

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                                  ---(2

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

-(α + β) = ba

: α + β = -ba

andαβ = cd

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) 2x² + 3x – 1 = 0

(b) 3x² – 5x – 2 = 0

(c) x² – 4x – 3 = 0

(d) ½ x² – 3x – 1 = 0

 

Solution

(a) 2x² + 3x – 1 = 0

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

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

α + β= -ba= -32

α β = ca= -12

(b) 3x² – 5x – 2 = 0

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

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

α + β = -ba= 53

α β = ca= -23

(c) x² – 4x – 3 = 0

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

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

α + β = -ba= 41

α β = ca= -3

(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 = -ba= (3)12=6

α β = ca= -112  = -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 = 12+ 5 = 112 ,   α β = 52

:The quadratic equation whose roots are 12 and  5  is

x²– 112 x + 52=  0

or 2x² – 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 =34+ 12= 54 ,   α β = 34 x 12= 38

:The quadratic equation whose roots are ¾ and ½ is

x²– 54 x + 38=  0

or 8x² – 10x + 3 = 0

Symmetric Properties of Roots

of ax² + bx + c = 0, then

α + β = -ba;   α β = ca

Certain relations involving α and β can also be determined from α + β and α β even when we do not knowα and β distinctively. Such relations are usually said to be symmetric.

They are symmetric in the sense that if α and β are interchanged, either the relation remains the same or is multiplied by -1.

If α≠ β, determine whether or not each of the following is symmetric:

(a) α + β                                   (b) αβ

(c) α² β²                                    (d) α²– β²

(e) 3α +2β                                (f) α² β²

 

Solution

(a) α+ β =β + α

: α + β is symmetric

 

(b)αβ = βα

: αβ is symmetric

 

(c) α² β²= α² β²                         

: α² β² is symmetric

 

(d) α² – β²= -(α² – β²)

: α² – β²is symmetric

 

(e) 3α + 2β≠ 3β + 2αsince α ≠ β

:3α + 2β is not symmetric                             

(f) α²+ β² = β²+α²

:α² + β²is symmetric

If α and β are the roots of 3x² – 4x – 1 = 0, find the value of:

(a) α+ β                                              (b) αβ 

(c) α² β²                                              (d) 1α+ 1β

(e)αβ+ βα                                               (f) α³β³

(g) α–β                                               (h)1α+1 + 1β+1

 

Solution

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

(a) α + β = - ba = 43                                               

Since D enables us to determine the position of the graph of y = ax² + bx + c relative to the x – axis, it is called a discriminant.

Determine the nature of roots of the following quadratic equations:

(i) x² – 3x – 2 = 0

(ii) x² – 6x + 9 = 0

(iii) 2x² – 2x + 5 = 0

 

Solution

(i) a = 1; b = -3; c = -2

D = b² – 4ac

= 9 + 8

= 17<0

Hence the roots of the equation are real and distinct.

(ii) x² – 2x + 9 = 0

a = 1; b = -6; c = 9

D = b² – 4ac

= 36 – 36

= 0

Hence the roots are real and equal.

(iii) 2x² – 2x + 5 = 0

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

D = b² – 4ac

= 4 – 40

= -36

Hence the roots are imaginary.

Evaluation

1. Find the quadratic equation where roots are

(a) 3 and -2            (b) ¾ and ½

 

General Evaluation

(1) If α and β are the roots of 3x² – 4x – 1 = 10, find the value of:

(a) α² + β²                                 (b) 1α+ 1β                           (c) αβ+ βα

(d) α³ + β³                                 (e) α – β

 

(2) Find the sum and product of the roots of these equations

(a) 2x² + 3x – 1 = 0                                      (b) 3x² – 5x – 2 = 0

Reading assignment

New Further Maths Project 2 pages 8, 9, 10, 11

Weekend Assignment

(1) Determine the nature of roots of x² – 3x – 2 = 0            

(a) Real                           (b) Imaginary                   (c) Equal                (d) Coincidental

(2) If α ≠ β which of the following is not symmetric

(a) αβ = βα                      (b) α + β = β + α             (c) 3α + 2β = 3β + 2α

(d) α² + β² = β² + α²

If α and β are the roots of 2x² – 7x – 3 = 0, find:

(3)αβ² + α²

(a)72b32c-214d421

(4) αβ +  βα

(a) -616b661c313d16

(5) 1α +  1β

(a) -37b-73c32d47

 

Theory

(1) Find the constants p, q and r such that 3x² – 12x + 16 = p (x + q)² + r

(2) If α and β are the roots of x² – 10x + 2 = 0, find α³ – β³.