Class 10 Maths Chapter 2 HOTS Questions - Polynomials

Q1: Find the Quadratic polynomial whose sum and product of zeros are √2 + 1, 1/√2 + 1 .
Ans: sum = 2 √2
Product = 1
Q.P = X² – (sum) x + Product
∴ x² – (2 √2 ) x + 1

Q2: If α,b are the zeros of the polynomial 2x² – 4x + 5 find the value of a) α² + β² b) (α - β)².
Ans: p (x) = 2 x² – 4 x + 5
α + β = -b/a = 4/2 = 2
αβ = c/a = 5/2
α² + β² = (α + β)² – 2αβ
Substitute then we get, α 2 + β2 = -1
(α - β)² = (α + β)² - 4 α β
Substitute, we get = (α - β)² = - 6

Q3: If the squared difference of the zeros of the quadratic polynomial x² + px + 45 is equal to 144 , find the value of p.
Ans: Let two zeros are a and b where α > β
According given condition
(α - β)² = 144
Let p(x) = x² + px + 45
α + β = -b/a = -p/1 = -p
αβ = c/a = 45/1 = 45
now (a - β)² = 144
(α + β)² – 4 αβ = 144
(-p)² – 4 (45) = 144
Solving this we get p = ± 18
Q4: If α, β are the zeros of a Quadratic polynomial such that α + β = 24, α - β = 8. Find a Quadratic polynomial having α and β as its zeros.
Ans: α + β = 24
α - β = 8
_________
2α = 32
α = 32/2 = 16, ∴ α = 16
Work the same way to α+β = 24
So, β = 8
Q.P is x² – (sum) x + product
= x² – (16+8) x + 16 x 8
Solve this,
it is k (x² – 24x + 128)
Q5: If α and β are the zeroes of the polynomial x² + 8x + 6 frame a quadratic polynomial whose zeroes are  
(a) 1/α and 1/β
(b) 1 + β/α, 1 + α/β.
Ans: Given polynomial x² + 8x + 6

Hence, the Required Quadratic polynomial f(x) is given by


Q6: If α & β are the zeroes of the polynomial 2x² - 4x + 5, then find the value of 
(i) α² + β²
(ii) 1/a + 1/ß 
(iii) (α - β)²
(iv) 1/α² + 1/β²
(v) α³ + β³
Ans: 



Q7: If the ratios of the polynomial ax³ + 3bx² + 3cx + d are in AP, Prove that 2b³ - 3abc + a²d = 0
Ans: Let the zeros of the given polynomial be p, q, r. As the roots are in A.P., then it can be assumed as p−k, p, p + k, where k is a common difference.

Q8: If α and β are the zeros of a quadratic polynomial such that  α+β=15 and  α−β=9, find the quadratic polynomial having α and β as its zeros.

Ans: Given:

α+β=15

α−β=9

From α+β=15:

2α = 15+9 = 24  ⇒ α=12

From α−β=9:

2β = 15−9 = 6  ⇒β=3

Quadratic polynomial:

$p(x)\; =\; x^2\; -\; (\backslash alpha\; +\; \backslash beta)x\; +\; \backslash alpha\backslash beta$p(x) = x²− (α+β)x +αβ

Substitute:

$p(x)\; =\; x^2\; -\; 15x\; +\; 36$p(x)=x²−15x+36