Q) While playing in a garden, Samaira saw a honeycomb and asked her mother what is that. Her mother replied that it’s a honeycomb made by honey bees to store honey. Also, she told her that the shape of the honeycomb formed is a mathematical structure. The mathematical representation of the honeycomb is shown in the graph.
Based on the above information, answer the following questions :
(i) How many zeroes are there for the polynomial represented by the graph given ?
(ii) Write the zeroes of the polynomial.
(iii) (a) If the zeroes of a polynomial x2 + (a + 1) x + b are 2 and – 3, then determine the values of a and b.
OR (iii) (b) If the square of difference of the zeroes of the polynomial x2 + px + 45 is 144, then find the value of p.
Ans:
(i) zeroes of the polynomial:
We know that the zeroes are represented based on no. of times, the graph cuts the X- axis.
Here in the given diagram, Polynomial graph cuts X-Axis 2 times,
Therefore there are 2 zeroes.
(ii) Zeroes of the polynomial:
Here, Polynomial cuts the X-axis at (7, 0) and (-7,0)
Hence, value of X are 7 and – 7
Therefore, zeroes of the polynomial are 7 and – 7.
(iii) (a) Value of a and B:
Polynomial given, f(x) = x 2 + (a + 1) x + b
By writing the given polynomial, f(x) = 0, we get:
x2 + (a + 1) x + b = 0
Next, comparing it with standard quadratic equation, A x 2 + B x + C = 0
we get, A = 1, B = (a + 1), C = b
We know that, Sum of the roots, α + β =
Since, Zeroes are given as 2, -3,
hence for given polynomial, Sum of the roots:
(2) + (- 3) =
∴ – 1 = – (a + 1)
∴ – 1 = – a – 1
∴ a = 0
Next, Product of the roots, α . β =
Hence, for given polynomial, product of the roots:
2 (- 3) =
∴ – 6 = b
∴ b = – 6
Therefore, values of a and b are 0 and – 6, respectively.
(iii) (b) Value of p:
Given the polynomial: f(x) = x2 + px + 45
By writing the given polynomial, f(x) = 0, we get:
x2 + p x + 45 = 0
Next, comparing it with standard quadratic equation, A x 2 + B x + C = 0
we get, A = 1, B = p, C = 45
We know that, Sum of the roots, α + β =
hence, if zeroes are α & β, then (α + β) =
∴ (α + β) = – p ……….. (i)
Next, Product of the roots, α . β =
Hence, for given polynomial, product of the roots:
α . β =
∴ α β = 45
∴ α = ……….. (ii)
Also, we are given that square of difference of the zeroes = 144
∴ (α – β) 2 = 144
∴ (α – β) 2 = (12)2
∴ (α – β) = 12 ………. (iii)
By substituting value of α from equation (ii) one by one, we get:
(α – β) = 12
∴ ( – β) = 12
∴ (45 – β2 ) = 12 β
∴ β2 + 12 β – 45 = 0
∴ β2 + 15 β – 3 β – 45 = 0
∴ β (β + 15) – 3 (β + 15) = 0
∴ (β + 15) (β – 3) = 0
∴ β = 3, β = – 15
Case 1, when β = 3,
α =
∴ α = View Post
∴ α = 15
Hence, α = 15 and β = 3
Now from equation (i),
(α + β) = – p
By substituting values of α & β, we get
15 + 3 = – p
∴ p = – 18
Case 2, when β = – 15,
α =
∴ α = – 3
Hence, α = – 3 and β = – 15
Now from equation (i),
(α + β) = – p
By substituting values of α & β, we get
– 3 + (- 15) = = p
∴ p = – 18
Therefore, in both cases, value of p is – 18.
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