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Polynomial Exercise 2.2

Hi Friends and champs!
Before we start next exercise, we need to know some basic terms:

Variables: A variable is a term which can have any value i.e. not constant (fix value), in general we use x, y, z ,..... etc.

Constant: A constant is a number that does not vary ex- 1,2,3,4......etc

Coefficient: A number which multiply or exist before a variable like 2 is a coefficient in 2x, 7 is a coefficient in 7x3. 

Polynomial: A polynomial is an expression, which include Variables, coefficient and constant like ax2+bx+c=0, polynomial could be monomial, binomial, trinomial......etc. 

Degree of a polynomial: is the highest power of variable, ex. in ax²+b degree is 1 and ax²+bx+c is of degree 2.

Zeroes of a polynomial: A real number "k" will be a ZERO of the polynomial p(x) if p(k)=0, i.e. on putting the constant k in place of "x" in the polynomial, the result is "0".

Now consider a quadratic polynomial p(x) such that:
p(x) = ax² + bx + c=0---------------(1)

let α,  β are the zeroes of the polynomial p(x).
divide by "a" in both side of equation (1)

x²+(b/a)x + c/a =0 ---------------(2)

since α,  β are the zeroes of the polynomial p(x), we can write eq (2) in form of

(x-α)(x-β) = 0

or x² - βx-αx + αβ = 0

x² - (α + β)x + αβ = 0 ------------(3) 

compare coefficient of equation (2) and (3) 
α + β = -b/a  ( i.e. sum of the zeroes is " -coefficient of x / coefficient of x² " )
αβ = c/a ( i.e. product of zeroes is " constant/coefficient of x² "  )

similarly, we can prove for degree 3 polynomial p(x)=ax3+bx2+cx+d=0, a ≠ 0
α + β + γ = -b/a
α β + β γ+γα = c/a
α β γ = -d/a


                                                                    Exercise -2.2
Q(1). Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. 
 (i) x² – 2x – 8          (ii) 4s² – 4s + 1        (iii) 6x² – 3 – 7x 
 (iv) 4u2 + 8u          (v) t2 – 15                  (vi) 3x2 – x – 4

Solution:
(i) x² – 2x – 8
= x²-4x+2x-8
=x(x-4)+2(x-4)
=(x-4)(x+2)
so zeroes are "4" and "-2"
sum of zeros =4+ (-2) =2
remember α + β = -b/a  ( i.e. sum of the zeroes is "-coefficient of x/coefficient of x²" )
therefore,
 α + β = -(-2/1)
=2 ( verified )
Product of Zeroes = 4 X -2 = -8
remember αβ = c/a ( i.e. product of zeroes is "constant/coefficient of x²"  )
therefore
αβ = -8/1
x
     = -8 ( verified )

(ii) 4s² – 4s + 1
4s² – 4s + 1 = 4s² - 2s-2s + 1
                   = 2s(2s-1)-1(2s-1)
                    =(2s-1)²
                           => s=1/2,1/2
sum of zeroes = 1/2 +  1/2 =1
                        = -(-4)/4 =1     ( -coefficient of s/ coefficient of s² )
Product of zeroes = 1/2 X 1/2 
                             =1/4           ( constant / coefficient of s² ) 
(iii) 6x² – 3 – 7x 
        6x² – 3 – 7x = 6x²  – 7x -3
                              = 6x² – 9x + 2x - 3
                              = 3x(2x-3) + 1(2x - 3)
                              = (3x + 1)( 2x - 3 )
So Zeroes are -1/3 and 3/2
sum of zeroes = -1/3 + 3/2
                        = (-2+9) /6
                        = 7/6 = ( - coefficient of x / coefficient of  x²) 
Product of zeroes = -1/3 X 3/2
                            = -1 / 2 
                            = -3/6 ( multiply and divide by 3 ) 
                            =  constant / coefficient of x²

(iv) 4u² + 8u    
        4u² + 8u  = 4u(u+2)
so zeroes are "0" and "-2"
sum of zeroes = 0 + (-2)
                        = -2
                        = -8/4 = ( - coefficient of x / coefficient of  x²) 
Product of zeroes = 0 X (-2 )
                            = 0
                            = 0/4  
                            =  constant / coefficient of x²                    

(v) t2 – 15
        t² – 15 = t² - (√15)²
                    = (t + √15 )(t - √15 )
so zeroes are -√15 and √15
sum of zeroes are = -√15 + √15
                               = 0 
                               = 0 / 1 ( - coefficient of t / coefficient of  t²)
product of zeroes = -√15 X √15
                            = -15 
                            = -15 / 1  =  constant / coefficient of x² 

(vi) 3x² – x – 4
                3x² – x – 4 = 3x² – 4x + 3x – 4
                                   = x(3x-4)+1(3x-4)
                                   =(x+1)(3x-4)
so zeroes are " -1 " and "4/3"
sum of zeroes = -1 + 4/3
                        = 1/3 = ( - coefficient of t / coefficient of  t²)
product of zeroes = -1 X 4/3
                            = -4/3  = constant / coefficient of x²

Q (2). Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i) 1/4 ,-1
Let the quadratic polynomial be ax² + bx + c, and its zeroes are  α and β 
then α + β = 1/4 = -b/a
and  αβ = -1
              = -1/1 = c/a
If a = 1 , b= -1/4 and c = -1
so equation is x² -(1/4)x -1
                    = 4x²-x-4

(ii) √2 , 1/3
then α + β = √2 = -b/a
and  αβ = 1/3
              = c/a
If a = 3, b= -3√2 and c = 1
so equation is 3x² - 3√2x + 1

(iii) 0 , √5 
then α + β = 0 = -b/a
and  αβ = √5
              = c/a
If a = 1, then b= 0 and c = √5
so equation is x²  + √5

(iv) 1 , 1
then α + β = 1 = -b/a
and  αβ = 1
              = c/a
If a = 1, then b= -1 and c = 1
so equation is x² - x + 1

(v) -1/4 , 1/4
then α + β = -1/4 = -b/a
and  αβ = 1/4 = c/a
If a = 4, then b= 1 and c = 1
so equation is 4x² + x + 1

(vi) 4 , 1
then α + β = 4 = -b/a
and  αβ = 1 = c/a
If a = 1, then b= -4 and c = 1
so equation is x² -4 x + 1


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