TOPICS
Exercise - 2.4

Question-1 :-  Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:
(i) 2x³ + x² - 5x + 2; 1/2, 1, -2
(ii) x³ – 4x² + 5x - 2; 2, 1, 1

Solution :-
(i) p(x) = 2x³ + x² - 5x + 2; 1/2, 1, -2 
    p(1/2) = 2 x (1/2)³ + (1/2)² - 5 x (1/2) + 2
           = 2 x 1/8 + 1/4 - 5/2 + 2
           = 1/4 + 1/4 - 5/2 + 2
           = (1 + 1 - 10 + 8)/4
           = 0/4 = 0
    p(1) = 2 x (1)³ + (1)² - 5 x 1 + 2
         = 2 + 1 - 5 + 2
         = 5 - 5 = 0 
    p(-2) = 2 x (-2)³ + (-2)² - 5 x (-2) + 2
         = 2 x (-8) + 4 + 10 + 2
         = -16 + 6 + 10
         = -16 + 16 = 0
   So, 1/2, 1, -2 are the zeroes of the cubic polynomial p(x) = 2x³ + x² - 5x + 2.

   In general, cubic polynomial is ax³ + bx² + cx + d
   p(x) = 2x³ + x² - 5x + 2
   By compairing, a = 2, b = 1, c = -5, d = 2.
   Relationship between the zeroes and the coefficients :
   Given that α = 1/2, β = 1, γ = -2
   α + β + γ = 1/2 + 1 - 2  
             = 3/2 - 2
             = -1/2  
        -b/a = -(1)/2 = -1/2
   αβ + βγ + γα = 1/2 x 1 + 1 x (-2) + (-2) x 1/2
                = 1/2 - 2  - 1 
                = (1 - 4 - 2)/2 
                = -5/2
            c/a = -5/2
   αβγ = 1/2 x 1 x (-2) = -1
  -d/a = -(2)/2 = -1
    
(ii) p(x) = x³ – 4x² + 5x - 2; 2, 1, 1 
    p(2) = (2)³ – 4 x (2)² + 5 x 2 - 2
         = 8 - 16 + 10 - 2
         = 18 - 18 = 0
    p(1) = (1)³ – 4 x (1)² + 5 x 1 - 2
         = 1 - 4 + 5 - 2
         = 6 - 6 = 0
   So, 2, 1, 1 are the zeroes of the cubic polynomial p(x) = x³ – 4x² + 5x - 2.

   In general, cubic polynomial is ax³ + bx² + cx + d
   p(x) = x³ – 4x² + 5x - 2
   By compairing, a = 1, b = -4, c = 5, d = -2.
   Relationship between the zeroes and the coefficients :
   Given that α = 2, β = 1, γ = 1
   α + β + γ = 2 + 1 + 1 = 4 
        -b/a = -(-4)/1 = 4
   αβ + βγ + γα = 2 x 1 + 1 x 1 + 1 x 2
                = 2 + 1 + 2 = 5
            c/a = 5/1 = 5
   αβγ = 2 x 1 x 1 = 2
  -d/a = -(-2)/1 = 2
    

Question-2 :-  Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively.

Solution :-
   In general, the cubic polynomial is ax³ + bx² + cx + d.
   Now, Given that α = 2, β = -7, γ = -14
   α + β + γ = 2 = -b/a = -(-2)/1
   αβ + βγ + γα = -7 = c/a = -(7)/1
   αβγ = -14 = -d/a = -(14)/1
   All coefficients a = 1, b = -2, c = 7, d = 14.
   Therefore, cubic polynomial is x³ - 2x² + 7x + 14.       
    

Question-3 :-  Obtain all other zeroes of If the zeroes of the polynomial x³ – 3x² + x + 1 are a – b, a, a + b, find a and b.

Solution :-
   p(x) = x³ – 3x² + x + 1
   Zeroes of polynomial are a - b, a, a + b.
   In general, the cubic polynomial is px³ + qx² + rx + s.
   So, p = 1, q = -3, r = 1, s = 1
   Sum of zeroes = a - b + a + a + b = 3a 
   -q/p = -(-3)/1 = 3
   3a = 3
    a = 1
   Product of zeroes = (a - b) x a x (a + b)
                     = a x (a² - b²)
                     = 1 x (1 - b²)
                     = 1 - b²
   -s/p = -(1)/1 = -1
    1 - b² = -1
      - b² = -1 - 1
      - b² = -2
        b² = 2
        b  = ±√2
   All other zeroes are 1, ±√2.
    

Question-4 :-  If two zeroes of the polynomial x⁴ – 6x³ – 26x² + 138x – 35 are 2 ± √3 , find other zeroes.

Solution :-
    p(x) = x⁴ – 6x³ – 26x² + 138x – 35
    Given that two zeroes are 2 + √3 and 2 - √3.
    Now, g(x) = (x - 2 - √3)(x - 2 + √3)
              = x² + 4 - 4x – 3
              = x² - 4x + 1
    Class 10 NCERT Math
    q(x) = x² - 2x - 35
         = x² - 7x + 5x -35
         = x(x - 7) + 5(x - 7)
         = (x + 5)(x - 7)
    x + 5 = 0
        x = -5
    x - 7 = 0
        x = 7 
    All zeroes of the polynomial are 2 + √3, 2 - √3, -5 and 7.
    

Question-5 :-  If the polynomial x⁴ – 6x³ + 16x² – 25x + 10 is divided by another polynomial x² – 2x + k, the remainder comes out to be x + a, find k and a.

Solution :-
    p(x) = x⁴ – 6x³ + 16x² – 25x + 10  
    g(x) = x² – 2x + k
    r(x) = x + a
    p(x) = g(x) x q(x) + r(x)
    p(x) - r(x) = g(x) x q(x)
    x⁴ – 6x³ + 16x² – 25x + 10 - x - a = (x² – 2x + k) x q(x)
    x⁴ – 6x³ + 16x² – 26x + 10 - a = (x² – 2x + k) x q(x)
    Class 10 NCERT Math
    q(x) = x² – 4x + (8 - k)
    r(x) = (-10 + 2k)x + (10 - a - 8k + k²) 
    -10 + 2k = 0
          2k = 10
           k = 5
    10 - a - 8k + k² = 0
    10 - a - 8 x 5 + 5² = 0
       10 - a - 40 + 25 = 0
                 -a - 5 = 0
                      a = -5
    a = -5, k = 5 
    
CLASSES

Connect with us:

Copyright © 2015-16 by a1classes.

www.000webhost.com