KPK Board 10th Class Math Ch 2 Theory of Quadratic Equation Short Questions Answers

KPK Board 10th Class Math Ch 2 Theory of Quadratic Equation Short Questions Answers

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For what value of K the roots of the equation

3x2- 5x+k=0 are equal.

Here                      a  =  3  ,   b  =  -5   ,  c = k

Since roots are equal then discriminate is equal to zero. i.e.

B2– 4ac=0 ………..(ii)put values in (ii)

=>           (-5)2 – 4 (3)(k) = 0                            =>              25   = 12k

=>            k = 25/12

Evaluate ( -1+ √(-3) )7+ ( -1- √(-3) )7.

Solution:              ( -1+ √(-3) )7+ ( -1- √(-3) )7

As                           (-1+i√3)/2 = w                          and                                 (-1+i√3)/2 =w2

=>           -1+i √3  =  2w                              and                          -1-i√3        =  2w2

Then

( -1+ √(-3) )7+ ( -1- √(-3) )7            =          (2w)7    +   (2w2 )7

=>           27 w7+27 w14

=>           27 (w2.w+w12.w2)

=>           27 [(w3 )2.w+(w3 )4.w2]

As w3   =  1

=>           2^7 ((1)3.w+(14.w2)

=>           27 (w+w2)

=>           128 (-1)                                 AS    w+ w2  = -1

=>           -128

With out solving the equation, find the sum and products of the roots of the following equation quadratic equation.

(i) 4x2- 1=0 (ii) 3x2+ 4x =0

Solution:              (i)            4x2+ 0.x –  1 =0

Here      a = 4       ,               b = 0      ,               c = -1

As sum of the roots   S = –  b/a……………(i)

Put values (i).

S = –  0/4                          =>      S = 0

Now product of the roots   =  p  =  S =   c/a  ……………..  (ii)      put values in ……(i)

=>           P = –  1/4

(ii)           3x2+ 4x+0 =  0

Here      a  =  3  ,                 b  =  4  ,                 c  =  0

As sum of the roots        S = –  b/a   …………(i)

Put values in (i).

S = –  4/3

Product of the roots  = S = –  c/a  put values

=>           P = –  0/3                                  =>                P  =  0

Find the value of k so that the sum of the roots of the equation 3x^2+ (2K+1 )x+k-5 = 0 is equal to the product of roots.

Solution:              3x2+ (2K+1 )x+k-5 =  0……………(i)

Here a  =  3   ,   b = 2k  + 1  ,  c   k  – 5

As sum of the roots  S = –  b/a  …………..(i)

Put values in (i).

=>               S = –  (2k+1)/3 …………………. (ii)

As product of the roots = P = –  c/a ………………… (iii)  put values in (iii)

P =   (k-5)/3…………….. (iv)

AS Sum of the roots = product of the roots

Then from (ii) and (iv)

–  (2k+1)/3  =    (k –  5)/3                              =>              -(2k + 1 ) = k – 5

=>           – 2k – 2 = k – 5                                    =>           – 2 + 5  = k + 2k

=>           3k   =   3                                                =>           K =   3/3 = 1

Find the value of k if the roots of x2- 3x+k+1=0 differ by unity.

Solution:              x2– 3x+k+1=0……………(i)

Let a and β are the roots of equation ……(i)

α-β =  1 ……………….. (i)

Here  a =  1  ,  b  =  -3  ,  c  =  k + 1

Sum of the roots                              α+β= -b/a

α+β-(-3)/1               =>           α+β=3 ……………………….(ii)

Product of the roots   αβ= c/a

αβ= (k+1)/1                                        =>           αβ= k+1…………….(iii)

Adding (i) and (ii)

α-β= 1

= (α+β=3)/(2a=4)

=>           α=4/2                     => α  =  2

Put α=2 in (ii)

2 + β = 3               =>        β  =3-2

=>           β = 1

Now put α=2  and  β =  1 in ( iii)

(2)(1) = k + 1        =>           2 – 1 = k

=>           k = 1

Thus k  = 1

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