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
α + β = -ba = -81 = -8
αβ = ca = 61 = 6
a) 1α + 1β = β + ααβ = -86 = -43
1α × 1β = 1αβ = 16
Hence, the Required Quadratic polynomial f(x) is given by
f(x) = k [x² − 43x + 16] = k (x² + 43x + 16), where k is a constant
b) (1 + βα) (1 + αβ) = 2 + βα + αβ
= 2αβ + β² + α²αβ = (α + β)²αβ = (-8)²6 = 646 = 323
Hence, the required quadratic polynomial f(x) is given by:
f(x) = k (x² + 323x + 323), where k is a constant

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: 
2x² − 4x + 5
α + β = −ba = −(−4)2 = 42 = 2
αβ = ca = 52
(α² + β²) = (α + β)² − 2αβ
= 4 − 5 = −1
1α + 1β = α + βαβ = 25
(α − β)² = (α² + β²) − 2αβ
= −1 − 2(52) = −1 − 5
= −6
(α³ + β³) = (α + β)((α² + β²) − αβ)
= (2)(−1 − 52)
= 2(−72)
= −7

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.
p + p + k = 3ba
⇒ p = −ba
And, (p − k)(p + k) = −da
(p² − k²) = −da
k² = db
Now, p(p − k) + p(p + k) + (p − k)(p + k) = 3ca
2p² + (p² − k²) = 3ca
2b²a² + db = 3ca
2b³ + a²da²b = 3ca
2b³ − 3abc + a²d = 0

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) = x²− (α+β)x +αβ
Substitute:
p(x)=x²−15x+36