For a linear polynomial kx + c, k ≠ 0, the graph  of y = kx + c is a straight line which intersects  the X-axis at exactly one point, namely, ((-c)/k,0), Therefore, the linear polynomial kx + c, k ≠ 0, has  exactly one zero, namely, the X-coordinate of the  point where the graph of y = kx + c intersects the  X-axis.

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Question 1

If a linear polynomial is 2x + 3, then the zero of  2x + 3 is:

(a) 3/2 

(b) − 3/2

(c) 2/3   

(d) − 2/3

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Question 2

The graph of y = p(x) is given in figure below for  some polynomial p(x). The number of zero/zeroes  of p(x) is/are:

(a) 1 

(b) 2

(c) 3    

(d) 0

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Question 3

If 𝛼 and 𝛽 are the zeroes of the quadratic polynomial  x 2 – 5x + k such that 𝛼 – 𝛽 = 1, then the value of k is:

(a) 4  

(b) 5

(c) 6  

(d) 3

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Question 4

If α and β are the zeroes of the quadratic polynomial  p(x) = 4x2 + 5x + 1, then the product of zeroes is:

(a) −1  

(b) 1/4

(c) −2  

(d) − 5/4

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Question 5

If the product of the zeroes of the quadratic  polynomial p(x) = ax 2 – 6x – 6 is 4, then the value  of a is:

(a) − 3/2  

(b) 3/2

(c) 2/3   

(d) − 2/3

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Transcript

Question For a linear polynomial kx + c, k β‰  0, the graph of y = kx + c is a straight line which intersects the X-axis at exactly one point, namely, ((βˆ’π‘)/π‘˜,0), Therefore, the linear polynomial kx + c, k β‰  0, has exactly one zero, namely, the X-coordinate of the point where the graph of y = kx + c intersects the X-axis. Give answer the following questions: Question 1 If a linear polynomial is 2x + 3, then the zero of 2x + 3 is: (a) 3/2 (b) βˆ’ 3/2 (c) 2/3 (d) βˆ’ 2/3 Let p(x) = 2x + 3 Finding zero p(x) = 0 2x + 3 = 0 2x = – 3 x = (βˆ’πŸ‘)/𝟐 So, the correct answer is (B) Question 2 The graph of y = p(x) is given in figure below for some polynomial p(x). The number of zero/zeroes of p(x) is/are: (a) 1 (b) 2 (c) 3 (d) 0 Number of zeroes is equal to number of times parabola intersects the x-axis Since the graph does not intersect the X-axis, ∴ Number of zeroes = 0 So, the correct answer is (d) Question 3 If 𝛼 and 𝛽 are the zeroes of the quadratic polynomial x2 – 5x + k such that 𝛼 – 𝛽 = 1, then the value of k is: (a) 4 (b) 5 (c) 6 (d) 3 Let p(x) = x2 – 5x + k Now, Sum of zeros = 𝒄/𝒂 𝛼 + 𝛽 = (βˆ’(βˆ’5))/1 𝛼 + 𝛽 = 5 Also given, 𝜢 βˆ’ 𝜷 = 1 Product of zeros = 𝒄/𝒂 𝛼𝛽 = π‘˜/1 𝛼𝛽 = k Adding (1) and (2) 𝛼 + 𝛽 + 𝛼 βˆ’ 𝛽 = 5 + 1 2𝛼 = 6 𝛼 = 6/2 𝛼 = 3 Putting 𝛼 = 3 in (1) 𝛼 + 𝛽 = 5 3 + 𝛽 = 5 𝛽 = 5 βˆ’ 3 𝛽 = 2 Now, from (3) 𝛼𝛽 = k 3 Γ— 2 = k 6 = k k = 6 So, the correct answer is (C) Question 4 If 𝛼 and 𝛽 are the zeroes of the quadratic polynomial p(x) = 4x2 + 5x + 1, then the product of zeroes is: (a) βˆ’1 (b) 1/4 (c) βˆ’2 (d) βˆ’ 5/4 Given p(x) = 4x2 + 5x + 1 Now, Product of Zeros = 𝑐/π‘Ž = 𝟏/πŸ’ So, the correct answer is (B) Question 5 If the product of the zeroes of the quadratic polynomial p(x) = ax2 – 6x – 6 is 4, then the value of a is: (a) βˆ’ 3/2 (b) 3/2 (c) 2/3 (d) βˆ’ 2/3 Given p(x) = ax2 – 6x – 6 Here, Product of zeroes = 𝑐/π‘Ž 4 = (βˆ’πŸ”)/𝒂 4a = βˆ’6 a = (βˆ’6)/4 a = (βˆ’πŸ‘)/𝟐 So, the correct answer is (A)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.