Case Based Questions: Polynomials | Mathematics (Maths) Class 10 PDF Download

Q1: Read the source below and answer the questions that follow:

Ramesh was asked by one of his friends Anirudh to find the polynomial whose zeroes are -2/√3 and √3/4.
He obtained the polynomial by following steps which are as shown below:
Let α = -2/√3 and β = √3/4
Then,
α + β = (-2/√3) + (√3/4) = (-8 + 1) / (4√3) = -7 / (4√3)
and
αβ = (-2/√3) × (√3/4) = -1/2
∴ Required polynomial = x² - (α + β)x + αβ
= x² - (-7/4√3)x + (-1/2)
= x² + (7x/4√3) - 1/2
= 4√3x² + 7x - 2√3
His another friend Kavita pointed out that the polynomial obtained is not correct.

i. Is the claim of Kavita correct? (1 mark)
ii. If given polynomial is incorrect, then find the correct quadratic polynomial. (1 mark)
iii. Find the value of α² + β². (2 mark)
Or
iii. If correct polynomial p(x) is a factor of (x - 2), then find f(2). (2 mark)

Ans:
i. Given, α = -2/√3 and β = √3/4
∴ α + β = (-2/√3) + (√3/4) = (-8 + 3) / 4√3 = -5 / 4√3
and αβ = (-2/√3) × (√3/4) = -1/2
Yes, because value of (α + β) calculated by Anirudh is incorrect.
ii. Required polynomial = k(x² - (α + β)x + αβ)
k(x² + (5x/4√3) - 1/2)
= (k/4√3)(4√3x² + 5x - 2√3)
= (4√3x² + 5x - 2√3) where k = 4√3
iii. α² + β² = (α + β)² - 2αβ
= (-5/4√3)² - 2 × (-1/2)
= 25/48 + 1 = 73/48
Alternate method:
α² + β² = (-2/√3)² + (√3/4)²
= 4/3 + 3/16 = 64/48 + 9/48 = 73/48
Or
iii. We have, p(x) = 4√3x² + 5x - 2√3
Since, p(x) is a factor of (x - 2), then
p(2) = 4√3(2)² + 5(2) - 2√3
= 16√3 + 10 - 2√3 = 14√3 + 10
Hence, remainder is 14√3 + 10.

Q2: Read the source below and answer the questions that follow:

Ans:
i. We have, f(x) = ax² + bx + c, a<0 It means a figure is a shape of parabola which open downwards.
ii. Since, x = 4 is one of the zeros of the polynomial (p - 1)x² + px + 1.
∴ (p - 1)(4)² + p(4) + 1 = 0
⇒ 16p - 16 + 4p + 1 = 0
⇒ 20p = 15
⇒ p = 3/4
iii. Required quadratic polynomial = k [x² - (Sum of zeroes)x + (Product of zeroes)] = k(x²-(-12+5)x+(-12)(5)) = k(x²+7x-60), where k is any arbitrary constant.  
iv. Let the zeroes of the quadratic polynomial be α. Since, one zero of the polynomial f(x) is reciprocal of the other. 
∴ Product of zeroes = (-1)² × (Constant term / Coefficient of x²)
⇒ α × (1/α) = 1 × (m/5) ⇒ m = 5

Q3: Read the source below and answer the questions that follow:

Ans:
i. The graph of a quadratic polynomial is a parabola which open upwards. 
ii. The zeroes of the quadratic polynomial p(x) = -8 -2x + x² are x-coordinates of the points where the graph intersects the X-axis. From the given graph, -2 and 4 are the x-coordinates of the points where the graph of p(x)=-8-2x+x² intersects the X-axis. Hence, -2 and 4 are zeroes of p(x)=-8-2x+x². 
iii. The graph of the given quadratic polynomial cut X-axis at points (-2, 0) and (4,0). 
iv. The graph of the given quadratic polynomial cut Y-axis at point (0, -8). 

Q4: Read the source below and answer the questions that follow:

i. Is Neha correct in her claim? (1 mark) 
ii. If the polynomial is incorrect, find the correct polynomial. (1 mark) 
iii. Find the value of α² + β². (1 marks)
iv. If the correct polynomial q(x) is a factor of (x+3), find q(-3). (1 marks)

Ans:
i. Yes, Neha is correct. 
The sum of roots should be α + β = 3/√2 + (-√2/5) = (15 - 2)/(5√2) = 13/(5√2), 
which is correct, but the final polynomial calculation by Pooja is incorrect.
ii. Correct Polynomial:
p(x) = k(x² - (α + β)x + αβ)
= k(x² - 13x/(5√2) - 3/5)
Multiplying by 5√2 to remove the denominator:
= 5√2x² - 13x - 3√2
where k = 5√2.
iii. Value of α² + β²:
α² + β² = (α + β)² - 2αβ
= (13/(5√2))² - 2(-3/5)
= 169/50 + 6/5
= 169+60/50 = 229/50.
iv. If q(x) = 5√2x² - 13x - 3√2 is a factor of (x + 3):
q(-3) = 5√2(-3)² - 13(-3) - 3√2
= 5√2(9) + 39 - 3√2
= 45√2 + 39 - 3√2
= 42√2 + 39.