MATHS Solutions For Class 10 MBOSE II Maths Ex 2(a)

 

CHAPTER-2             



     POLYNOMIALS


EXERCISE- 2(a).


SHORT ANSWER TYPE QUESTION (SLAB-I)


1.  FIND THE ZEROES OF THE POLYNOMIAL.

(i). x2-2x-3

SOLUTION:   LET,   P(x) = x2-2x-3

             ∴ P(x)= x2-2x-3

              = x2-3x+x-3

             = x(x-3)+1(x-3)

             = (x-3)(x+1)

NOW,

          P(x) = 0

(x-3) = 0                               OR,                        (x+1) = 0

x = 3                                    0R,                         x=-1

 

THE ZEROES OF P(x) = (3,-1).

 

(ii). 49x2-64


SOLUTION: LET,  P(x) = 49x2-64

∴ P(x)= 49x2-64

       = (7x)2-82.

       = 7x – 8

∴ P(x)= 0

x= -8/7        OR,      X = 8/7.

THE ZEROES ARE (-8/7, 8/7)

 

 

 

 

(iii). 3x2-4-4x


SOLUTION:  LET,   P(x) = 3x2-4x-4

              P(x) = 3x2-(6-2)x-4

                            = 3x2-6x+2x-4

                            = 3x(x-2)+2(x+2)

                            = (x-2) (3x+2)

 

NOW,

                    P(x) = 0

          x-2 = 0                                    OR,                          3x+2  = 0

              x = 2                                    OR,                                 x = -2/3

 

THE ZEROES ARE  (2 , -2/3).

 

(iv).  4√3x2 + 5x + 23


SOLUTION:  LET,   P(x) = 43x2 + 5x - 23

                            = 43x2 + 8x – 3x - 23

                            = 4x(3x+2) -  3(3+2)

                            = (4x-3)  (3+2)

 

          NOW,

                    4x - 3 = 0                         OR,              (3+2) = 0

                    x = 3/4                              OR,                        x = -2/3


THE ZEROES ARE (3/4 , -2/3).

 

3. IF THE ZEROES OF THE QUADRATIC POLYNOMIALS P(x) = 3x2+(2K-1)x-5 ARE EQUAL IN MAGNITUDE BUT OPPOSITE IN SIGN THEN FIND THE VALUE OF K.


SOLUTION:           P(x) = 3x2+(2x-1)-5

                    LET, THE ZEROES BE α AND β.

A/Q ,

          α = -α


SUM OF THE ROOTS = α+ β = 0

                                        = - (2K-1)/3           = 0

                                        = (-2K + 1)/3         = 0

                                        = -2K + 1                = 0

                                        ⟹ K                     = -1/-2

                          K = 1/2


4. IF ONE ZERO THE POLYNOMIAL (a2+9)x2+15x+6a IS RECIPROCAL OF THE OTHER, FIND THE VALUE OF a.


SOLUTION:  IF ONE ZERO IS THE RECIPROCAL OF THE OTHER,

          THEN,          LET ONE ZERO BE ‘a’

OTHER ZERO = 1/a

          NOW,

                    PRODUCT OF ZEROES = c/a

                                      ⟹ a + 1/a                    = 6/a2+9

                                               1                   = 6a/a2+9

                                         a2 + 9                   = 6a

                                     ⟹ a2 + 9 – 6a              = 0

                                     ⟹ a2 – 6a + 9              = 0

                                        ⟹ a2 - 3a – 3a + 9        = 0

                                          a(a-3) – 3(a-3)        = 0

                                        ⟹ (a-3) (a-3)                 = 0

          a = 3.

5. IF ONE ZERO OF THE POLYNOMIAL q(x) = x2 + px + 8 IS DOUBLE THE OFFER,THEN FIND THE VLAUE OF p AND HENCE BOTH ZEROES OF q(x).


SOLUTION:           LET THE ZEROES BE α AND β.

A/Q ,

          α  = 2β

NOW,

                        α + β = -b/a

                  2β + β = -p

                       3β = -p

                         β = -p/3                                    ⟶ (i)         

           

AGAIN,

                              α * β    = c/a

  2β * β     = 8

  2          = 8

  β2           = 4

  (-p/3)2   = 4                               ⟶ [ FROM EQ. i]

  p2/9        = 4

  p2            = 36

  p              = ±√36

  p              = ±6.


PUTTING THE VALUE OF P IN Eq. (i), WE GET

         

            β      = -p/3                           Or,                         β        = - (-p/3)

                     = -6/3                                                                   = -(-6/3)

            β = -2.                                                                               = -(-2)

                                                                                        β   = 2

 

 

    α+β      =-b/a                               Or,                                 α+β      = -b/a

= α+(-2) =-p                                                                    α+2     = -p

= α-2        =-6                                                                 ⟹ α+2      = -(-6)

= α            =-6+2                                                                      ⟹ α            =6-2

= α            =-4                                                                 α             = 4.

 

      THE ZEROES ARE 2, 4 WHEN P = -6 AND THE ZEROES ARE -2 , -4. WHEN P = 6.

 

6. FIND K   IF ZEROES α , β OF THE POLYNIMIAL 5x2+(2k+7)x + (k-2) ARE SUCH THAT 2α + 5β = 1.

SOLUTION:  GIVEN,

                              α AND  β ARE THE ZEROES OF THE POLYNOMIAL

P(x) = 5x2 + (2k+1)x + (k-2)

NOW,

                  α + β       = -b/a

            α + β        = -(2k+1)/5

           ⟹ 5α + 5β    = -2k – 1                                                                    ⟶ (i)

AND,

                              α ⌿ β =  c/a

                        ⟹ α ⌿ β = (k – 2)/5                                                ⟶(ii)

GIVEN,

                    2α + 5β = 2

              ⟹ 5β = (1 - 2α)                                                                   ⟶ (iii)

          PUTTING Eq. (iii) IN Eq. (i)

WE GET,

                    5α + 1 - 2α = -2k-1

                              = -2k – 2                                                   

                   α               = -2k-2/3                                                   ⟶(iv)

PUTTING Eq. (iv) IN Eq (iii),

                                       = 1 – 2(-2k-2)/3

                                  = 3 + 4k + 4 /15

                          β         =   4k + 17/15                                         ⟶(v)

NOW,

          FROM Eq (ii),(iv) AND (v),

WE GET,

                    (-2k-2)/3 * 4k + 7/15                     = k-2/5

          ⟹ (- 8k2 - 14K – 8k – 14)/3*3                = k – 2   

          ⟹ -8K2 – 22k – 14                                       = 9k – 18

          ⟹ -8k2 – 31k + 4 – 9k + 18                      = 0

           ⟹ 8k2 + 32k – k – 4                                    = 0

          ⟹ 8k (k+4) – 1(k+4)                                  = 0

          ⟹ (8k-1) (k+1)                                              = 0

  

      k = 1/8                                     Or,                             k = -4.

 

7. FIND A QUADRATIC POLYNOMIAL EACH WITH THE NUMBERS GIVEN BELOW AS SUM AND PRODUCT OF ITS ZEROES RESPECTEVILY.

          (i). -3/5, -2/5        (ii). 8/3, 7             (iii). (23 – 1) , (3 - 3).


(i). -3/5 , -2/5

SOLUTION:    GIVEN,

                    α + β = -3/5       = -b/a

          AND,

                    α + β = -2/5       = c/a

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

          QUADRATIC POLYNOMIAL         = ax2 + bx + c

                                                                                      = 5X2 + 3x – 2.


(ii). 8/3 , 7

SOLUTION: GIVEN,

                    α  + β = 8/b                 = -b/a

AND

          α . β   = 7                                  = c/a

  a = 3                     b = -8                    c = 7

QUADRATIC POLYNOMIAL                   = ax2 + bx + c

                                                                             = 3x2 – 8x + 7

 

(iii). (23 – 1), (3 - √3)

SOLUTION:   GIVEN,

                              α + β = 2√3 – 1           = -b/a

AND,

          α . β = 3 - √3 = c/a

 

QUADRATIC POLYNOMIAL = ax2 + bx + c.

                                                           = x2 – (23-1) + (3-3).       





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