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NCERT solution for class 10th maths: chapter-2 POLYNOMIALS (Ex.2.2)


Question-1: Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
Solution: (i)  x² - 2x - 8

 x² -2-8=0
 x² -(4-2)x-8=0   
 x² -4x+2x-8=0       
(x² -4x)+(2x-8) =0
x(x-4) + 2(x-4) =0
(x+2) (x-4)=0
these are the factors of the equation.But we have to find the value of x.

(x+2) = 0     and  (x-4)=0
x = -2                    x= 4
Therefore zeroes of polynomial (x²-2x-8) are 4 and -2.
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Now, second part of question is verify the relationship.

Sum of zeroes of given equation =(-2+4) = 2





Now,
Product of zeroes of given equation = (-2 x 4) = -8






 Hence the relationship is verified.
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(ii) 4s² - 4s +1

4s² - 4s +1 = 4s² -4s +1
                  = 4s² -(2+2)s +1
                  = 4s² -2s-2s +1
                  = (4s² -2s) - (2s +1)    
                  = 2s(2s-1) -1(2s-1)      
                  = (2s-1) (2s-1)
 So, factors of equation are (2s-1), (2s-1)
now find the value of s
                  2s-1= 0                      2s-1= 0
                   2s= 1                            2s=1
                     s= 1/2                           s= 1/2
                                                         
So, zeroes of polynomial (4s² -4s +1) are  1/2 and   1/2
                                                               
Now, verification








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Now,

 Hence the relationship is verified.
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(iii) 6x² -3 -7x 

6x² -7x -3         
= 6x² -(9-2)x -3
= 6x² -9x+2x -3
= (6x² -9x)+(2x -3)
= 3x(2-3) +1(2x -3)  
= (3x +1) (2x -3)         
Now, find the value of
 3x +1=0                                   2x -3= 0
     3x = -1                                    2x 3
       x = -1/3                                   x = 3/2 
So,zeroes of polynomial (6x² -7x -3) are -1/3 and 3/2.

Now, verification






--------------------------------------------------
Now,







 Hence the relationship is verified.
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(iv) 4u²+8u

4u²+8u = 4u(u+2)
so, here factors of this equation are 4u and (u+2)
Now, we will find the value of u.
4u =0           u+2=0
  u=0/4         u= -2
  u=0        
So, zeroes of polynomial (4u²+8u) are 0 and -2.

Now, verification
                     











Hence the relationship is verified.
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(v) t² - 15

t² - (√15)²

 t² - (√15)² = (t-√15)(t+√15)  
Now find the value of t
t-√15 = 0                         t+√15 = 0
t = √15                             t = -√15
So,zeroes of polynomial (t² - 15) are √15 and -√15 .

Now, verification

Sum of zeroes of given equation = -√15 +√15  = 0


Now,





Hence the relationship is verified.
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(vi) 3x² -x -4

3x²-x-4 = 3x² -(4-3)-4
             = 3x² -4x+3-4
             = x(3x-4)+1(3x-4) 
             = (x+1)(3x-4)

Now find the value of x
x+1 = 0                      3x -4 = 0
    x = -1                     3x = 4
                                    x = 4/3
So,zeroes of polynomial (3x² --4) are -1 and 4/3.

Now, verification












Hence the relationship is verified.
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Question-2: Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
Solution: (i)  1/4 , -1

ax²+bx+c=0
First Method        
  ax²+bx+c=0
     






Now,
Sum of zeroes = -b/a = -1/4
Product of zeroes = c/a = -4/4
means a= 4, b=-1 and c=-4
 ax²+bx+c
Quadratic polynomial =  4x²-1x-4 or  4x² --4

Second method with formula 
Sum of zeroes (𝛂+𝜷) = 1/4
Product of zeroes (𝛂 x 𝜷) = -1
formula for equation = x²-(𝛂+𝜷)x+(𝛂 x 𝜷)   
                                  = x²- +(-1)
                                           4

So, quadratic equation is 4x² --4.

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(ii)2, 1/3

Sum of zeroes (𝛂+𝜷) = 2
Product of zeroes (𝛂 x 𝜷) = 1/3
formula for equation = x² -(𝛂+𝜷)+(𝛂 x 𝜷)
                                   = x²-(2)x+ 1/

 So, quadratic equation is 3x² - 32x +1

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(iii)0, 5                              
 Sum of zeroes (𝛂+𝜷) = 0
Product of zeroes (𝛂 x 𝜷) = √5
formula for equation = x² -(𝛂+𝜷)+(𝛂 x 𝜷)
                                  =x² -(0)x√5
                                  x² -0 x x + √5
So, quadratic equation is  x² + 5
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(iv) 1, 1
 Sum of zeroes (𝛂+𝜷) = 1
Product of zeroes (𝛂 x 𝜷) = 1
formula for equation = x² -(𝛂+𝜷)+(𝛂 x 𝜷)
                                  =x² -1x+ 1
                                  x² -x1
So, quadratic equation is x² -+1
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(v) -1/4, 1/4
Sum of zeroes (𝛂+𝜷) = -1/4
Product of zeroes (𝛂 x 𝜷) = 1/4
                               formula for equation = x² -(𝛂+𝜷)+(𝛂 x 𝜷)

So, quadratic equation is 4x² +x +1.

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(vi) 4, 1
Sum of zeroes (𝛂+𝜷) = 4
Product of zeroes (𝛂 x 𝜷) = 1
formula for equation = x² -(𝛂+𝜷)+(𝛂 x 𝜷)
                                  =x² -4x1
                                 
So, quadratic equation is x² -4x +1
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NCERT solution for class 10th maths: chapter-2 POLYNOMIALS (Ex.2.1)


Question-1: The graphs of y= p(x) are given in the figure (A) as shown below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.
Solution:

                                                                                           












(i) The graph is parallel to x-axis. It does not intersect the x-axis. So the number of zeroes is 0.

NCERT solution for class 10th maths: chapter-1 REAL NUMBERS (Ex.1.4)

                                       
Question-1: Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or non-terminating repeating decimal expansion.
Solution:   



factors of 3125 = 5×5×5×5×5



So, here 3125 can be written in the form of 5ⁿ, therefore we can say that this fraction
(13/3125)  has terminating decimal expansion.
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 Factors of 8 = 2×2×2



So, here 8 can be written in the form of 2³, therefore we can say that this fraction (17/8) has terminating decimal expansion.
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Factors of 455 = 5×7×13



So, here 455 cannot be written in the form of
therefore we can say that this fraction (64/455) has Non-terminating repeating decimal expansion.
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Factors of 1600 = 2×2×2×2×2×2×5×5




So, here 1600 can be written in the form of  
therefore we can say that this fraction (15/1600) has terminating decimal expansion. 
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  Factors of 343 = 7×7×7
                                                                                                   

So, here 343 cannot be written in the form of
therefore we can say that this fraction (29 /343 ) has Non-terminating repeating decimal expansion.  
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So, here 2³5² is in the form of
therefore we can say that this fraction (23/2³5²) has terminating decimal expansion.
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So, here 2²5⁷7⁵ is not in the form of
therefore we can say that this fraction (129/2²5⁷7⁵) has Non-terminating repeating decimal expansion.
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  Factors of 15 = 3×5
                                                                                                 



So, here denominator 5 can be written in the form of  5ⁿ, therefore we can say that this fraction (6/15) has terminating decimal expansion.
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Factors of 50 = 2×5×5



So, here 50 can be written in the form of 
therefore we can say that this fraction (35/50) has terminating decimal expansion.
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  Factors of 210 = 2×3×5×7
                                                                                                  



So, here 210 cannot be written in the form of 
therefore we can say that this fraction (77/210) has Non-terminating repeating decimal expansion. 
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Question-2: Write down the decimal expansions of those rational numbers in question 1 above which have terminating decimal expansions.
Solution:  
                                                                                                                                
                                                                                                                                       

                      
                                                                                                                                                                                                                                                                                 *********************************************************************************
         
                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                 

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Question-2: The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and the form of p/q, what can you say about the prime factors of q? 
(i) 43.123456789
(ii) 0.120120012000120000....



Solution:  

(i) 43.123456789
it has certain number of digits so they can be represented in form of p/q.
Hence they are rational number.  

As they have certain number of digit and the number which has certain number of digits is always terminating number  and for terminating number denominator has prime factor 2 and 5 only .



(ii) 0.120120012000120000. . . 
In this problem repetitions number are not same so it is a irrational number.  



In this number 0.123456789 repeating again and again so it is a rational number and it is none terminating so that the prime factor has a value which is not equal to 2 or 5.