Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

Q1: Find the sum and product of zeroes of 3x² - 5x + 6.

Here, p (x) = 3x² - 5x + 6
Comparing it with ax² + bx + c, we have
a = 3, b = - 5, c = 6

Sum of the zeroes:

= - Coefficient of x / Coefficient of x² = - (-5)/3 = 53

Product of the zeroes:

= Constant term / Coefficient of x² = 6/3 = 2

Q2: Find the sum and product of the zeroes of polynomial p (x) = 2x³ - 5x_² - 14x + 8.

Comparing p (x) = 2x³ - 5x² - 14x + 8 with ax³ + bx² + cx + d, we have
a = 2, b = –5,
c = - 14 and d = 8

Sum of the zeroes:

= -b/a = -(-5)/2 = 5/2

Product of the zeroes:

= -d/a = -8/2 = -4

Q3: Find a quadratic polynomial whose zeroes are: 2 + √5 / 2 and 2 - √5 / 2 .

Sum of zeroes (S):
(2 + √5)/2 + (2 - √5)/2 = 4/2 = 2
Product of zeroes (P):

(2 + √5)/2 × (2 - √5)/2 = [2² - (√5)²]/4 = 4 - 5/4 = -1/4
Quadratic Polynomial:
The required quadratic polynomial is:
k(x² - Sx + P), where k is any real number.
Substituting the values of S and P:
k(x² - 2x + (-1/4)) = k(x² - 2x - 1/4)

Thus, the required polynomial is
= k (x² - 2x - 1/4)

Q4: If α and β are the zeroes of a Quadratic polynomial x² + x - 2 then find the value of .

Comparing x² + x - 2 with ax² + bx + c, we have:
a = 1, b = 1, c = - 2

Sum of roots (α + β) = -ba = -(1)/1 = -1

Product of roots (αβ) = ca = (-2)/1 = -2

To find:

1/α - 1/β

Solution:

1/α + 1/β = β - α/αβ = -(-3)/-2 = 3/2

Now, calculate (α - β):

(α - β)² = (α + β)² - 4αβ
(α - β)² = 1² - 4(-2)
(α - β)² = 1 + 8 = 9
(α - β) = ±√9 = ±3

Therefore:
1/α - 1/β = - 3/2

Q5: If a and b are the zeroes of x² + px + q then find the value of ( αβ + 2) ( βα + 2).

Comparing x² + px + q with ax² + bx + c
a =1, b = p and c = q
∴ Sum of zeroes, a + b = - b/a
⇒
and αβ = c/a
⇒ αβ = q/1 = q
Now,

= αβ + 2αβ + 2βα + 4

Substituting α + β = -p and αβ = q:
= 5 + 2[ (α + β)² - 2αβαβ]

Further simplifying:
= 5 + 2[ p² - 2qq]

Final Result:

The value of the given expression is:
( αβ + 2) ( βα + 2) = 2p² + qq

Q6: Find the zeroes of the quadratic polynomial 6x² - 3 - 7x.

We have,
= 6x² - 3 - 7x = 6x² - 7x - 3
= 6x² - 9x + 2x - 3
= 3x (2x - 3) + 1 (2x - 3)
= (3x + 1) (2x - 3)
For 6x² - 3 - 7x to be equal to zero,
either (3x + 1) = 0 or (2x - 3) = 0
⇒ 3x = - 1 or 2x = 3
⇒
Thus, the zeroes of and 3/2.

Q7: Find the zeroes of 2x² - 8x + 6.

We have,
2x² - 8x + 6 = 2x² - 6x - 2x + 6
= 2x (x - 3) - 2 (x - 3)
= (2x - 2) (x - 3)
= 2 (x - 1) (x - 3)
For 2x² - 8x + 6 to be zero,
Either, x - 1 = 0 ⇒ x = 1
or x - 3 = 0 ⇒ x = 3
∴ The zeroes of 2x² - 8x + 6 are 1 and 3.

Q8: Find the zeroes of the quadratic polynomial 3x² + 5x - 2.

We have,
p (x) = 3x² + 5x - 2
= 3x² + 6x - x - 2
= 3x (x + 2) - 1 (x + 2)
= (x + 2) (3x - 1)
For p (x) = 0, we get
Either x + 2 = 0 ⇒ x = - 2
or 3x - 1 = 0 ⇒ x = 1/3
Thus, the zeroes of 3x² + 5x - 2 are - 2 and 1/3.

Q9: If the zero of a polynomial p (x) = 3x² - px + 2 and g (x) = 4x² - q x - 10 is 2, then find the value of p and q.

∵ p (x) = 3x² - px + 2
∴ p (2) = 3 (2)² - p (2) + 2 = 0
[2 is a zero of p (x)]
or 12 - 2p + 2 or 14 - 2p = 0
or p = 7
Next g (x) = 4x² - q x - 10
∴ g (2) = 4(2)² - Q (2) - 10 = 0
[2 is a zero of g (x)]
or 4 × 4 - 2q - 10 = 0
or 16 - 2q - 10 = 0
or 6 - 2q = 0
⇒ q = 6/2 ⇒ q = 3
Thus, the required values are p = 7 and q = 3.

Q10: Find the value of ‘k’ such that the quadratic polynomial 3x² + 2kx + x - k - 5 has the sum of zeroes as half of their product.

Here, p (x) = 3x² + 2kx + x - k - 5
= 3x² + (2k + 1) x - (k + 5)
Comparing p (x) with ax² + bx + c, we have:
a = 3, b = (2k + 1),
c = - (k + 5)
∴ Sum of the zeroes

Product of the zeroes

According to the condition,
Sum of zeroes = 1/2 (product of roots)

⇒ - 2 (2k + 1) = - (k + 5)
⇒ 2 (2k + 1) = k + 5
⇒ 4k + 2 = k + 5
⇒ 4k - k = 5 - 2
⇒ 3k = 3
⇒ k = 3/3 = 1

Q11: Find the zeroes of the polynomial f (x) = 2 - x².

We have f (x)= 2 - x²
= (√2 )² - x²

Q12: Find the cubic polynomial whose zeroes are 5, 3 and - 2.

∵ 5, 3 and - 2 are zeroes of p (x)
∴ (x - 5), (x - 3) and (x + 2) are the factors of p (x)
⇒ p (x) = k (x - 5) (x - 3) (x + 2)
= k (x² - 8x + 15) (x + 2)
= k (x³ - 8x² + 15x + 2x² - 16x + 30
= k (x³ + [- 8 + 2] x² + [15 - 16] x + 30)
= k (x³ - 6x² - x + 30)
Thus, the required polynomial is k (x³ - 6x² - x + 30).

Q13: If α, β and γ be the zeroes of a polynomial p (x) such that (α + β +γ) = 3, (αβ + βγ + γα) = -10 and αβγ = - 24 then find p (x).

Here, α + β + γ = 3
αb + βγ + γα = - 10
αβγ = - 24
∵ A cubic polynomial having zeroes as α,β,γ is
p (x) = x³ - (a + b + γ) x² + (αβ + βγ + γα) x - (αβγ)
∴The required cubic polynomial is
= k {x³ - (3) x² + (- 10) x - (- 24)}
= k(x³ - 3x² - 10x + 24)

Note: If α, β and γ be the zeroes of a cubic polynomial p (x) then
p (x) = x³ - [Sum of the zeroes] x² + [Product of the zeroes taken two at a time] x - [Product of zeroes]
i.e., p (x) = k {x³ - (α + β + γ) x² + [αβ + βγ + γα] x - (αβγ).

Q14: If α and β are the zeroes of the quadratic polynomial p (x) = kx² + 4x + 4 such that α² + β² = 24, find the value of k.

Here, p (x) = kx² + 4x + 4.
Comparing it with ax² + bx + c, we have:
a = k; b = 4; c = 4
∴ Sum of the zeroes = -b/a
⇒ α + β = -4/k
and Product of the zeroes = c/a
⇒ αβ = 4/k
∵ α² + β² = 24
∴ (α + β)² - 2αβ = 24
[∵ (x + y)² = x² + y² + 2xy ⇒ (x + y)² - 2xy = x² + y²]
⇒
⇒
⇒ 16 − 8k − 24k² =0
⇒ 24k² + 8k − 16 = 0
⇒ (3k − 2) (k + 1) = 0
⇒ 3k − 2= 0 or k + 1 = 0
⇒ k = 2/3 or k = -1

Q15: Find the zeroes of the quadratic polynomial 6x² - 3 - 7x and verify the relationship between the zeroes and the coefficients of the polynomial.

Here, p (x) = 6x² - 3 - 7x = 6x² - 7x - 3
= 6x² - 9x + 2x - 3
= 3x (2x - 3) + 1 (2x - 3)
= (2x - 3) (3x + 1)
=
∴ Zeroes of p (x) are 3/2 and
To verify the relationship:
Sum of the zeroes = - coefficient of x/coefficient of x²

⇒
⇒
⇒ 7/6 = 7/6
L.H.S = R.H.S ⇒ Relationship is verified.
Product of the zeroes = Constant term/coefficient of x²

⇒
⇒
i.e., L.H.S = R.H.S ⇒ Relationship is verified.

Q16: Find the zeroes of the quadratic polynomial 5x² - 4 - 8x and verify the relationship between the zeroes and the coefficients of the polynomial.

p (x) = 5x² - 4 - 8x
= 5x² - 8x - 4
= 5x² - 10x + 2x - 4
= 5x (x - 2) + 2 (x - 2)
= (x - 2) (5x + 2)

∴ zeroes of p (x) are 2 and
Relationship Verification
Sum of the zeroes = - coefficient of x/coefficient of x²
⇒
⇒
⇒ 8/5 = 8/5

i.e., L.H.S. = R.H.S. ⇒ relationship is verified.
Product of the zeroes = Constant term/coefficient of x²
⇒
⇒
i.e., L.H.S. = R.H.S.
⇒ The relationship is verified.

Q17: Find the quadratic polynomial, the sum of whose zeroes is 8 and their product is 12. Hence, find the zeroes of the polynomial.

The quadratic polynomial p (x) is given by
x² - (Sum of the zeroes) x + (Product of the zeroes)
∴ The required polynomial is
= x² - [8] x + [12]
= x² - 8x + 12
To find zeroes:
∵ x² - 8x + 12 = x² - 6x - 2x + 12
= x (x - 6) - 2 (x - 6)
= (x - 6) (x - 2)
∴ The zeroes of p (x) are 6 and 2.

Q18: If one zero of the polynomial (a² - 9) x² + 13x + 6a is reciprocal of the other, find the value of ‘a’.

Here, p (x) = (a² - 9) x² + 13x + 6a
Comparing it with Ax² + Bx + C, we have:
A = (a² - 9); B = 13; C = 6
Let one of the zeroes = a
∴ The other zero = 1/α
Now, Product of the zeroes

⇒
⇒ 6a = a² − 9    ⇒ a² − 6a + 9 = 0
⇒ (a − 3)² =0 ⇒ a − 3=0
⇒ a = 3
Thus, the required value of a is 3.

Q19: If the product of zeroes of the polynomial ax² - 6x - 6 is 4, find the value of ‘a’

Here, p (x) = ax² - 6x - 6
∵ Product of zeroes = Constant term/coefficient of x²
but product of zeroes is given as 4
∴  ⇒ − 6 = 4 × a
⇒  ⇒

Thus, the required value of a is -3/2.

Q20: Find the quadratic polynomial whose zeroes are 1 and - 3. Verify the relation between the coefficients and the zeroes of the polynomial.

The given zeroes are 1 and - 3.
∴ Sum of the zeroes = 1 + (- 3) = - 2
Product of the zeroes = 1 × (- 3) = - 3
A quadratic polynomial p (x) is given by
x² - (sum of the zeroes) x + (product of the zeroes)
∴ The required polynomial is
x² - (- 2) x + (- 3)
⇒ x² + 2x - 3
Verification of relationship
∵ Sum of the zeroes Constant term/coefficient of x²

∴
⇒− 2= − 2

i.e., L.H.S = R.H.S ⇒ The sum of zeroes is verified
∵ Product of the zeroes = Constant term/coefficient of x²
∴
⇒− 3= − 3

i.e., L.H.S = R.H.S ⇒ The product of zeroes is verified.

Q21: Find the zeroes of the quadratic polynomial 4x² - 4x - 3 and verify the relation between the zeroes and its coefficients.

Here, p (x) = 4x² - 4x - 3 = 4x² - 6x + 2x - 3
= 2x (2x - 3) + 1 (2x - 3)
= (2x - 3) (2x + 1)
=
∴  are zeroes of p (x).

Verification of relationship
∵ Sum of the zeroes = - coefficient of x/coefficient of x²

∴ ⇒ 3 - 1  2
⇒ 2/2 = 1 ⇒ 1= 1
⇒
2/2 = 1 ⇒ 1 = 1
i.e., L.H.S = R.H.S ⇒ Sum of zeroes is verified
Now, Product of zeroes =Constant term/coefficient of x²

⇒  
i.e., L.H.S = R.H.S ⇒ Product of zeroes is verified.

Q22: Find a quadratic polynomial whose zeroes are - 4 and 3 and verify the relationship between the zeroes and the coefficients.

We know that:
P (x) = x² - [Sum of the zeroes] x + [Product of the zeroes] ...(1)
∵ The given zeroes are - 4 and 3
∴ Sum of the zeroes = (- 4) + 3 = - 1
Product of the zeroes = (- 4) × 3 = - 12
From (1), we have
x² - (- 1) x + (- 12)
= x² + x - 12 ...(2)
Comparing (2) with ax² + bx + c, we have
a = 1, b = 1, c = - 12
∴ Sum of the zeroes = -b/a
⇒ (+ 3) + (- 4) = -1/1
i.e., L.H.S = R.H.S ⇒ Sum of zeroes is verified.
Product of zeroes = c/a
⇒ 3 × (- 4) = -12/1
⇒ - 12 = - 12
i.e., L.H.S = R.H.S ⇒ Product of roots is verified.

Q23: Find the zeroes of the polynomial x² + (1/6)x - 2 and verify the relation between the coefficients and the zeroes of the above polynomial.

The given polynomial is x² + (1/6)x - 2

Step 1: Write as:
= 6x² + x - 12/6

Step 2: Expand:
= 6x² + 9x - 8x - 12/6

Step 3: Factorize:
= 3x(2x + 3) - 4(2x + 3)/6

Step 4: Take common factors:
= (3x - 4)(2x + 3)/6

∴ zeroes of the given polynomial are 4/3  and - 3/2
Now in, x² + 16x - 2 
co-efficient of x² = 1
co-efficient of x = 1/6
constant term = –2
Sum of zeroes = 4/3 + -3/2 = 1/6 = coefficient of x/coefficient of x²

Product of zeroes = 4/3 × -3/2 = -2 = constant term/coefficient of x²

Q24: Find the quadratic polynomial, the sum and product of whose zeroes are √2 and -3/2 respectively. Also, find its zeroes.

Sum of zeroes = √2

Product of zeroes = -3/2

∴ A quadratic polynomial is given by:
x² - [sum of roots]x + [product of roots]
∴ The required polynomial is:
x² - √2x + -(3/2)

Since:

x² - √2x - (3/2) = (2x² - 2√2x - 3)/2

= 1 [2x² + √2x - 3(2x - 3)]/2
= 1 [(√2x)(√2x + 1) - 3(√2x - 3)]/2
= 1 [(√2x + 1)(√2x - 3)]/2

⇒ Zeroes are -1/√2 and 3/√2

Q25: If a and b are zeroes of the quadratic polynomial x² – 6x + a; find the value of ‘a’ if 3α + 2β = 20.

We have quadratic polynomial = x² – 6x + a ...(1)
∵ a and b are zeroes of (1)
∴

It is given that: 3α + 2β = 20      ...(2)
Now, α +β = 6    ⇒ 2 (α+ β) = 2(6)
2α + 2β = 12      ...(3)
Subtracting (3) from (2), we have

Substituting a = 8 in α + β= 6, we get
8 +β = 6 ⇒ β = –2
Since, αβ = a
8(–2) = α ⇒ α = –16