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CBSE Solutions for Class 10 English

GSEB std 10 science solution for Gujarati check Subject Chapters Wise::

If the sum of zeroes of the quadratic polynomial 3x2 – kx + 6 is 3, then find the value of k. (2012)

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જવાબ : Here p = 3, q = -k, r = 6
Sum of the zeroes, (α + β) = -q/p= 3 …..(given)
 \frac { -(-k) }{ 3 } = 3
k = 9


If α and β are the zeroes of the polynomial ay2 + by + c, find the value of α2 + β2.

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જવાબ : Important Questions for Class 10 Maths Chapter 2 Polynomials Q2


If the sum of the zeroes of the polynomial p(x) = (k– 14) x2 – 2x – 12 is 1, then find the value of k. (2017 D)

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જવાબ : p(x) = (k2 – 14) x2 – 2x – 12
Here a = k2 – 14, b = -2, c = -12
Sum of the zeroes, (α + β) = 1 …[Given in question]
⇒ \frac { -b }{ a } = 1
⇒ \frac { -\left( -2 \right) }{ { k }^{ 2 }-14 } = 1
⇒ k2 – 14 = 2
⇒ k2 = 16
⇒ k = ±4


If α and β are the zeroes of a polynomial such that α + β = -6 and αβ = 5, then find the polynomial. (2016 D)

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જવાબ : Quadratic polynomial is x2 – (Sum)x + (Product) = 0
⇒ x2 – (-6)x + 5 = 0
⇒ x2 + 6x + 5 = 0


Question 5.
A quadratic polynomial, whose zeroes are -4 and -5, is …. (2016 D)

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જવાબ : x2 + 9x + 20 is the required polynomial.


Find the condition that zeroes of polynomial p(x) = ax2 + bx + c are reciprocal of each other. (2017 OD)

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જવાબ : Let α and \frac { 1 }{ \alpha } be the zeroes of P(x).
P(a) = ax2 + bx + c …(given in the question)
Product of zeroes = \frac { c }{ a }
⇒ α × \frac { 1 }{ \alpha } = \frac { c }{ a }
⇒ 1 = \frac { c }{ a }
⇒ a = c
Coefficient of x2 = Constant term


Form a quadratic polynomial whose zeroes are 3 + √2 and 3 – √2. (2012)

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જવાબ : Sum of zeroes = (3 + √2) + (3 – √2) = 6
Product of zeroes = (3 + √2) x (3 – √2) = (3)2 – (√2)2 = 9 – 2 = 7
Quadratic polynomial = x2 – 6x + 7


Find a quadratic polynomial, the stun and product of whose zeroes are √3 and \frac { 1 }{ \surd 3 } respectively. (2014)

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જવાબ : Sum of zeroes = √3
Product of zeroes = \frac { 1 }{ \surd 3 }
Quadratic polynomial =
Important Questions for Class 10 Maths Chapter 2 Polynomials Q8


Find a quadratic polynomial, the sum and product of whose zeroes are 0 and -√2 respectively. (2015)

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જવાબ : Quadratic polynomial is
x2 – (Sum of zeroes) x + (Product of zeroes)
= x2 – (0)x + (-√2)
= x2 – √2


Find the zeroes of the quadratic polynomial √3 x2 – 8x + 4√3. (2013)
 

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જવાબ : Important Questions for Class 10 Maths Chapter 2 Polynomials Q10


If the zeroes of the polynomial x2 + px + q are double in value to the zeroes of 2x2 – 5x – 3, find the value of p and q. (2012)

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જવાબ : We have, 2x2 – 5x – 3 = 0
= 2x2 – 6x + x – 3
= 2x(x – 3) + 1(x – 3)
= (x – 3) (2x + 1)
Zeroes are:
x – 3 = 0 or 2x + 1 = 0
⇒ x = 3 or x = \frac { -1 }{ 2 }
Since the zeroes of required polynomial is double of given polynomial.
Zeroes of the required polynomial are:
3 × 2, (\frac { -1 }{ 2 } × 2), i.e., 6, -1
Sum of zeroes, S = 6 + (-1) = 5
Product of zeroes, P = 6 × (-1) = -6
Quadratic polynomial is x2 – Sx + P
⇒ x2 – 5x – 6 …(a)
Comparing (a) with x2 + px + q
p = -5, q = -6


Can (x – 2) be the remainder on division of a polynomial p(x) by (2x + 3)? Justify your answer. (2016 OD)

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જવાબ : In case of division of a polynomial by another polynomial, the degree of the remainder (polynomial) is always less than that of the divisor. (x – 2) can not be the remainder when p(x) is divided by (2x + 3) as the degree is the same.


Find a quadratic polynomial whose zeroes are \frac { 3+\surd 5 }{ 5 } and \frac { 3-\surd 5 }{ 5 }. (2013)

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જવાબ : Important Questions for Class 10 Maths Chapter 2 Polynomials Q13               = a(25x2 – 30x + 4 )                                           … where [a є R


Find the quadratic polynomial whose zeroes are -2 and -5. Verify the relationship between zeroes and coefficients of the polynomial. (2013)

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જવાબ : Sum of zeroes = (-2) + (-5) = -7
Product of zeroes = (-2)(-5) = 10
Quadratic polynomial = x2 – (-7)x + 10
= x2 + 7x + 10
Verification:
Here a = 1, b = 7, c = 10
Sum of zeroes = (-2) + (-5) = 7
Important Questions for Class 10 Maths Chapter 2 Polynomials Q14
Important Questions for Class 10 Maths Chapter 2 Polynomials Q14.1


Find the zeroes of the quadratic polynomial 3x2 – 75 and verify the relationship between the zeroes and the coefficients. (2014)

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જવાબ : We have, 3x2 – 75
= 3(x2 – 25)
= 3(x2 – 52)
= 3(x – 5)(x + 5)
Zeroes are:
x – 5 = 0 or x + 5 = 0
x = 5 or x = -5 ------(1)
Verification:
Here a = 3, b = 0, c = -75
Sum of the zeroes = 5 + (-5) = 0       [from eq.1]
Important Questions for Class 10 Maths Chapter 2 Polynomials Q15


Find the zeroes of p(x) = 2x2 – x – 6 and verify the relationship of zeroes with these co-efficients. (2017 OD)

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જવાબ : p(x) = 2x2 – x – 6 …[Given in question]
= 2x2 – 4x + 3x – 6
= 2x (x – 2) + 3 (x – 2)
= (x – 2) (2x + 3)
Zeroes are:
x – 2 = 0 or 2x + 3 = 0
x = 2 or x = \frac { -3 }{ 2 }
Verification:
Here a = 2, b = -1, c = -6
Important Questions for Class 10 Maths Chapter 2 Polynomials Q16


What must be subtracted from the polynomial f(x) = x4 + 2x3 – 13x2 – 12x + 21 so that the resulting polynomial is exactly divisible by x2 – 4x + 3? (2012, 2017 D)

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જવાબ : Important Questions for Class 10 Maths Chapter 2 Polynomials Q17
(2x – 3) should be subtracted from x4 + 2x3 – 13x2 – 12x + 21 as it is the remainder


Find a quadratic polynomial, the sum and product of whose zeroes are -8 and 12 respectively. Hence find the zeroes. (2014)

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જવાબ : Let Sum of zeroes (α + β) = S = -8 …[Given in question]
Product of zeroes (αβ) = P = 12 …[Given in question]
Quadratic polynomial is x2 – Sx + P
= x2 – (-8)x + 12
= x2 + 8x + 12
= x2 + 6x + 2x + 12
= x(x + 6) + 2(x + 6)
= (x + 2)(x + 6)
Zeroes are:
x + 2 = 0 or x + 6 = 0
x = -2 or x = -6


Find a quadratic polynomial, the sum and product of whose zeroes are 0 and \frac { -3 }{ 5 } respectively. Hence find the zeroes. (2015)

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જવાબ : Quadratic polynomial = x2 – Sx + P
Important Questions for Class 10 Maths Chapter 2 Polynomials Q21​​​​​​​


Divide 3x2 + 5x – 1 by x + 2 and verify the division algorithm. (2013 OD)

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જવાબ : Important Questions for Class 10 Maths Chapter 2 Polynomials Q25
Remainder = 1
Quotient = (3x – 1)
Verification:
Divisor × Quotient + Remainder
= (x + 2) × (3x – 1) + 1
= 3x– x + 6x – 2 + 1
= 3x2 + 5x – 1
= Dividend


If sum of the squares of zeroes of the quadratic polynomial 6x2 + x + l is 25/36, the value of l is

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જવાબ : -2


The value of k for which the polynomial x3 + 4x2 –kx + 8 is exactly divisible by (x – 2) is

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જવાબ : 16


If one of the zeroes of the quadratic polynomial (k – 1)x2 + mx + 1 is –3, then the value of m is

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જવાબ : 4/3


If one of the zeroes of the cubic polynomial x3 + ax2 + bx + c is –1, then the

Product of the other two zeroes is

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જવાબ : b – a + 1


Given that one of the zeroes of the cubic polynomial ax3 + bx2 + cx + d is zero, the product of the other two zeroes is

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જવાબ : c/a


If the zeroes of the quadratic polynomial x2 + (a + 1)x + b are 2 and –3, then a&b ?

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જવાબ : a = 0, b = – 6


The degree of the polynomial (x + 1)(x2 – x – x4 +1) is:

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જવાબ : 3


The number of polynomials having zeroes as –2 and 5 is ______________

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જવાબ : Infinite


Sign of the zeroes of the quadratic polynomial x2 + 99x + 127 are

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જવાબ : Both Negative


If (x - 2) and [x - ½ ] are the factors of the polynomials qx2 + 5x + r prove that q = r

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જવાબ : 4q+10+r=0 -----(a)
q/4 +5/2 +r=0 or q+10+4r=0 -----(b)
Subtracting a from b
3q-3r=0
q=r


If the zeroes of the quadratic polynomial ax2 + bx + c, c ≠ 0 are equal, then which constant should have same signs?

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જવાબ : a & c


A cubic polynomial is given below
S(x) =x3 -3x2+x+1
The zeroes of the polynomial are given as (a-b) ,a  and (a+b). What is the value a and b

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જવાબ : a=1,b=√2 or -√2


p(x) = x4 -6x3 +16x2 -25x +10
q(x) = x2-2x+k
It is given
p(x) = r(x) q(x) + (x+m)
Find the value of k and m

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જવાબ : K=5 and a=-5


If the zeroes of the quadratic equation are 11 and 2 ,what is expression for quadratic?

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જવાબ : (x-11)(x-2)


Z(x) = px2+(p-2)x +2. If  2 is the zero of this polynomial,what is the value of p

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જવાબ : -1/2


If p and q are the zeroes of the polynomial  x2-11x +30, Find the value of p3 + b3

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જવાબ : 341


Find the remainder  when x4+x3-2x2+x is divided by x-1

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જવાબ : 1


Factorize the following : 1+8x3

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જવાબ : (2x+1)(4x2-2x+1)


Factorize the following : x3-23x2+142x-120

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જવાબ : (x-1)(x-10)(x-12)


Factorize the following : 3x3 -x2-3x+1

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જવાબ : (3x-1)(x-1)(x+1)


Factorize the following : x2 +9x+18

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જવાબ : (x+6)(x+3)


State whether the statement is true or  false :- Graph of constant polynomial never meets x axis

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જવાબ : True


State whether the statement is true or  false :- Graph of polynomial (x2-1) meets the  x-axis at one point

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જવાબ : False


State whether the statement is true or  false :- The degree of zero polynomial is not defined

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જવાબ : True


State whether the statement is true or  false :- 1,2 are the zeroes of x2-3x+2

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જવાબ : True


State whether the statement is true or  false :- A binomial may have degree 6

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જવાબ : True


State whether the statement is true or  false :- Every real number is the zero’s of zero polynomial

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જવાબ : True


State whether the statement is true or  false :- Every linear polynomial has only one zero

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જવાબ : True


State whether the statement is true or  false :- The factor of 3x2 -x-4 are  (x+1)(3x-4)

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જવાબ : True


State whether the statement is true or  false :- P(x) =x-1 and g(x) =x2-2x +1 .  p(x) is a factor of g(x)

 

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જવાબ : True


Divide 4x3 + 2x2 + 5x – 6 by 2x2 + 1 + 3x and verify the division algorithm. (2013)

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જવાબ : Important Questions for Class 10 Maths Chapter 2 Polynomials Q28

Divisor [D]= (2x2 + 3x + 1)

Quotient [Q] = (2x – 2)
Remainder [R] = (9x – 4)
Verification:
[D] × [Q] + [R]
= (2x2 + 3x + 1) × (2x – 2) + 9x – 4
= 4x3 – 4x2 + 6x2 – 6x + 2x – 2 + 9x – 4
= 4x3 + 2x2 + 5x – 6
= Dividend


Given that x – √5 is a factor of the polynomial x3 – 3√5 x2 – 5x + 15√5, find all the zeroes of the polynomial. (2012, 2016)

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જવાબ : Let,the polynomial  P(x) = x3 – 3√5 x2 – 5x + 15√5
x – √5 is a factor of the given polynomial.
Put x = -√5,
Important Questions for Class 10 Maths Chapter 2 Polynomials Q29
Another zero:
x – 3√5 = 0 x = 3√5
All the zeroes of P(x) are -√5, √5 and 3√5.


If a polynomial x4 + 5x3 + 4x2 – 10x – 12 has two zeroes as -2 and -3, then find the other zeroes. (2014)

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જવાબ : Given two zeroes are -2 and -3.
(x + 2)(x + 3) = x2 + 3x + 2x + 6 = x2 + 5x + 6
Dividing the given equation with x2 + 5x + 6, we get
Important Questions for Class 10 Maths Chapter 2 Polynomials Q30
x4 + 5x+ 4x2 – 10x – 12
= (x2 + 5x + 6)(x2 – 2)
= (x + 2)(x + 3)(x – √2 )(x + √2 )
Other zeroes are:
x – √2 = 0 or x + √2 = 0
x = √2 or x = -√2


Find all the zeroes of the polynomial 8x4 + 8x3 – 18x2 – 20x – 5, if it is given that two of its zeroes are \sqrt { \frac { 5 }{ 2 } } and -\sqrt { \frac { 5 }{ 2 } }. (2014, 2016 D)

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જવાબ : Important Questions for Class 10 Maths Chapter 2 Polynomials Q31
Important Questions for Class 10 Maths Chapter 2 Polynomials Q31.1


If p(x) = x3 – 2x2 + mx + 5 is divided by (x – 2), the remainder is 11. Find m. Hence find all the zeroes of x3 + mx2 + 3x + 1.

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જવાબ : p(x) = x3 – 2x2 + mx + 5,
When x – 2,
p(2) = (2)3 – 2(2)2 + m(2) + 5
11 = 8 – 8 + 2m + 5
11 – 5 = 2m
6 = 2m
m = 3
Let q(x) = x3 + mx2 + 3x + 1
= x3 + 3x2 + 3x + 1
= x3 + 1 + 3x2 + 3x
= (x)3 + (1)3 + 3x(x + 1)
= (x + 1)3
= (x + 1) (x + 1) (x + 1) …[ a3 + b3 + 3ab (a + b) = (a + b)3]
All zeroes are:
x + 1 = 0 x = -1
x + 1 = 0 x = -1
x + 1 = 0 x = -1
Hence zeroes are -1, -1 and -1.


If α and β are zeroes of p(x) = kx2 + 4x + 4, such that α2 + β2 = 24, find k. (2013)

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જવાબ : We have, p(x) = kx2 + 4x + 4
Here a = k, b = 4, c = 4
Important Questions for Class 10 Maths Chapter 2 Polynomials Q33
24k2 = 16 – 8k
24k2 + 8k – 16 = 0
3k2 + k – 2 = 0 …[Dividing equation by 8]
3k2 + 3k – 2k – 2 = 0
3k(k + 1) – 2(k + 1) = 0
(k + 1)(3k – 2) = 0
k + 1 = 0 or 3k – 2 = 0
k = -1 or k = \frac { 2 }{ 3 }


If α and β are the zeroes of the polynomial p(x) = 2x2 + 5x + k, satisfying the relation, α2 + β2 + αβ = \frac { 21 }{ 4 } then find the value of k. (2017 OD)

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જવાબ : Given , p(x) = 2x2 + 5x + k
Here a = 2, b = 5, c = k
Important Questions for Class 10 Maths Chapter 2 Polynomials Q34


What must be subtracted from p(x) = 8x4 + 14x3 – 2x2 + 8x – 12 so that 4x2 + 3x – 2 is factor of p(x

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જવાબ : Important Questions for Class 10 Maths Chapter 2 Polynomials Q35
Remainder = (15x – 14).

Hence, Polynomial to be subtracted by (15x – 14).


Find the values of a and b so that x4 + x3 + 8x2 +ax – b is divisible by x2 + 1. (2015)

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જવાબ : Important Questions for Class 10 Maths Chapter 2 Polynomials Q36
If x4 + x3 + 8x2 + ax – b is divisible by x2 + 1
=> Remainder = 0
(a – 1)x – b – 7 = 0
(a – 1)x + (-b – 7) = 0 . x + 0
a – 1 = 0, -b – 7 = 0
a = 1, b = -7
a = 1, b = -7


If a polynomial 3x4 – 4x3 – 16x2 + 15x + 14 is divided by another polynomial x2 – 4, the remainder comes out to be px + q. Find the value of p and q. (2014)

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જવાબ : Important Questions for Class 10 Maths Chapter 2 Polynomials Q37


If the polynomial (x4 + 2x3 + 8x2 + 12x + 18) is divided by another polynomial (x2 + 5), the remainder comes out to be (ax +b ), find the values of a and b.

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જવાબ : Important Questions for Class 10 Maths Chapter 2 Polynomials Q38
Remainder = 2x + 3
ax + b = 2x + 3
a = 2 and b = 3


Find the condition that zeroes of polynomial p(x) = ax2 + bx + c are reciprocal of each other. (2017 OD)

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જવાબ : Let s and 1/ s be the zeroes of P(x).
P(x) = ax2 + bx + c …(given in question)
Product of zeroes = c/a
s × 1/s = c/a
1 = c/a

a = c (Required condition)
Coefficient of x2 = Constant term


Form a quadratic polynomial whose zeroes are 3 + √2 and 3 – √2. (2012)

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જવાબ : Sum of zeroes = (3 + √2) + (3 – √2) = 6
Product of zeroes = (3 + √2) x (3 – √2) = (3)2 – (√2)2 = 9 – 2 = 7
Quadratic polynomial = x2 – 6x + 7


Find a quadratic polynomial, the sum and product of whose zeroes are √3 and 1/√3 respectively. (2014)

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જવાબ : Sum of zeroes = √3
Product of zeroes = 1/√3

Quadratic polynomial = x2 – Sx + P
Important Questions for Class 10 Maths Chapter 2 Polynomials Q8


Find the zeroes of p(x) = 2x2 – x – 6 and verify the relationship of zeroes with these co-efficients. (2017 OD)

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જવાબ : p(x) = 2x2 – x – 6 …[Given in question]
= 2x2 – 4x + 3x – 6
= 2x (x – 2) + 3 (x – 2)
= (x – 2) (2x + 3)
Zeroes are:
x – 2 = 0 or 2x + 3 = 0
x = 2 or x = -3 / 2  
Verification:
Here a = 2, b = -1, c = -6
Important Questions for Class 10 Maths Chapter 2 Polynomials Q16​​​​​​​


If the zeroes of the polynomial x2 + px + q are double in value to the zeroes of 2x2 – 5x – 3, find the value of p and q. (2012)

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જવાબ : We have, 2x2 – 5x – 3 = 0
= 2x2 – 6x + x – 3
= 2x(x – 3) + 1(x – 3)
= (x – 3) (2x + 1)
Zeroes are:
x – 3 = 0 or 2x + 1 = 0
x = 3 or x =  -1 /2

Since the zeroes of required polynomial is double of given polynomial.
Zeroes of the required polynomial are:
3 × 2, -1/2× 2, i.e., 6, -1
Sum of zeroes, S = 6 + (-1) = 5
Product of zeroes, P = 6 × (-1) = -6
Quadratic polynomial is x2 – Sx + P
x2 – 5x – 6 …(A)
Comparing (A) with x2 + px + q
p = -5, q = -6


Find the zeroes of the quadratic polynomial √3 x2 – 8x + 4√3. (2013)

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જવાબ : Given equation, √3 x2 – 8x + 4√3=0
Important Questions for Class 10 Maths Chapter 2 Polynomials Q10


Find a quadratic polynomial whose zeroes are (3+5)/5  & (3-5)/5. (2013)

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જવાબ : Important Questions for Class 10 Maths Chapter 2 Polynomials Q13


Find the quadratic polynomial whose zeroes are -2 and -5. Verify the relationship between zeroes and coefficients of the polynomial. (2013)

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જવાબ : Sum of zeroes = (-2) + (-5) = -7
Product of zeroes = (-2)(-5) = 10
Quadratic polynomial is x2 – Sx + P = 0
= x2 – (-7)x + 10
= x2 + 7x + 10
Verification:
Here a = 1, b = 7, c = 10
Sum of zeroes = (-2) + (-5) = 7
Important Questions for Class 10 Maths Chapter 2 Polynomials Q14
Important Questions for Class 10 Maths Chapter 2 Polynomials Q14.1​​​​​​​


Find the zeroes of the quadratic polynomial 3x2 – 75 and verify the relationship between the zeroes and the coefficients. (2014)

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જવાબ : 3x2 – 75
= 3(x2 – 25)
= 3(x2 – 52)
= 3(x – 5)(x + 5)
Zeroes are x – 5 = 0 or x + 5 = 0
x = 5 or x = -5
Verification:
Here a = 3, b = 0, c = -75
Sum of the zeroes = 5 + (-5) = 0
Important Questions for Class 10 Maths Chapter 2 Polynomials Q15​​​​​​​


There are No Content Availble For this Chapter

Degree of polynomial

1

 x5 +x -25

A

0

2

(x+1)(x3 + x2 + 31)

B

2

3

9

C

5

4

x2+1

D

4

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જવાબ :

1-C, 2-D, 3-A, 4-B

Equations and Root

1

X2+ 10x + 25

A

-5 & 5

2

X2-5x-50

B

5 & 5

3

X2 -25

C

-2 & 20

4

X2+18x-40

D

-10 & 5

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જવાબ :

1-B, 2-D, 3-A, 4-C

Equations and Root

1

X2+ 31x + 30

A

-20 & -8

2

X2-28x+160

B

-7 & 7

3

X2 -49

C

2 & 3

4

X2+5x+ 6

D

1 & 30

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જવાબ :

1-D, 2-A, 3-B, 4-C

Degree of polynomial

1

( X2 +x -25) / (x+1)

A

6

2

(x+1)(x5 + x2 + 31)

B

1

3

X49/x7

C

10

4

(x2+1)5

D

42

 

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જવાબ :

1-B, 2-A, 3-D, 4-C

Equations and Root

1

X2 + 16x + 60

A

-20 & 4

2

X2 + 16x – 80

B

6 & 10

3

X2 -16x -80

C

8 & 10

4

X2 + 18x + 80

D

20 & -4

 

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જવાબ :

1-B, 2-D, 3-A, 4-C

P(x)=5x3 -3x2+7x+1

1

 P(1)

A

1

2

P(0)

B

10

3

P(-1)

C

-65

4

P(-2)

D

-14

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જવાબ :

1-B, 2-A, 3-D,4-D

Equations and Root

1

X2 + 12x + 36

A

-6 & -6

2

X2 + 12x – 160

B

6 & 6

3

X2 -12x + 36

C

-14 & 2

4

X2 -12x – 28

D

20 & -8

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જવાબ :

1-B, 2-D, 3-A, 4-C

Equations and Root

1

X2 + 17x + 30

A

18 & -1

2

X2 + 17x – 18

B

15 & 2

3

X2 -17x + 16

C

-14 & -3

4

X2 -17x + 42

D

-16 & -1

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જવાબ :

1-B, 2-A, 3-D, 4-C

Equations and Root

1

X2 + 11x + 10

A

-12 & 1

2

X2 + 11x + 18

B

-13 & 2

3

X2 -11x -12

C

10 & 1

4

X2 -11x - 26

D

9 & 2

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જવાબ :

1-C, 2-D, 3-A, 4-B

Equations and Root

1

X2 + 10x + 25

A

15 & -5

2

X2 + 10x - 45

B

5 & 5

3

X2 -10x - 96

C

-14 & -3

4

X2 -10x + 25

D

-5 & -5

 

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જવાબ :

1-B, 2-A, 3-D,4-D

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Polynomials

Chapter 02 : Polynomials

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