# NCERT Solutions for Class 10th: Ch 2 Polynomials Maths

#### NCERT Solutions for Class 10th: Ch 2 Polynomials Maths

Page No: 28**Exercises 2.1**

1. The graphs of y = p(x) are given in following figure, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

**Answer**

*x*-axis at any point.

(ii) The number of zeroes is 1 as the graph intersects the

*x*-axis at only 1 point.

(iii) The number of zeroes is 3 as the graph intersects the

*x*-axis at 3 points.

(iv) The number of zeroes is 2 as the graph intersects the

*x*-axis at 2 points.

(v) The number of zeroes is 4 as the graph intersects the

*x*-axis at 4 points.

(vi) The number of zeroes is 3 as the graph intersects the

*x*-axis at 3 points.

Page No: 33

**Exercise 2.2**

(i)

*x*

^{2}– 2

*x*– 8

(ii) 4

*s*

^{2}– 4

*s*+ 1

(iii) 6

*x*

^{2}– 3 – 7

*x*

(iv) 4

*u*

^{2}+ 8

*u*

(v)

*t*^{2}– 15
(vi) 3

*x*^{2}–*x*– 4**Answer**

(i)

*x*

^{2}– 2

*x*– 8

= (

*x*- 4) (

*x*+ 2)

The value of

*x*

^{2}– 2

*x*– 8 is zero when

*x*- 4 = 0 or

*x*+ 2 = 0, i.e., when

*x*= 4 or

*x*= -2

Therefore, the zeroes of

*x*

^{2}– 2

*x*– 8 are 4 and -2.

Sum of zeroes = 4 + (-2) = 2 = -(-2)/1 = -(Coefficient of

*x*)/Coefficient of*x*^{2}
Product of zeroes = 4 × (-2) = -8 = -8/1 = Constant term/Coefficient of

*x*^{2}(ii) 4

*s*

^{2}– 4

*s*+ 1

= (2

*s*-1)

^{2}

The value of 4

*s*

^{2}- 4

*s*+ 1 is zero when 2

*s*- 1 = 0, i.e., s = 1/2

Therefore, the zeroes of 4s

^{2}- 4s + 1 are 1/2 and 1/2.Sum of zeroes = 1/2 + 1/2 = 1 = -(-4)/4 = -(Coefficient of

*s)*/Coefficient of

*s*

^{2}

Product of zeroes = 1/2 × 1/2 = 1/4 = Constant term/Coefficient of

*s*

^{2}.

(iii) 6

*x*

^{2}– 3 – 7

*x*

*=*6

*x*

^{2 }– 7

*x*– 3

= (3

*x*+ 1) (2

*x*- 3)

The value of 6

*x*

^{2}- 3 - 7

*x*is zero when 3

*x*+ 1 = 0 or 2

*x*- 3 = 0, i.e.,

*x*= -1/3 or

*x*= 3/2

Therefore, the zeroes of 6

*x*^{2}- 3 - 7*x*are -1/3 and 3/2.Sum of zeroes = -1/3 + 3/2 = 7/6 = -(-7)/6 = -(Coefficient of

*x*)/Coefficient of

*x*

^{2}

Product of zeroes = -1/3 × 3/2 = -1/2 = -3/6 = Constant term/Coefficient of

*x*

^{2}.

(iv) 4

*u*

^{2}+ 8

*u*

*=*4

*u*

^{2}+ 8

*u +*0

= 4

*u*(

*u*+ 2)

The value of 4

*u*

^{2}+ 8

*u*is zero when 4

*u*= 0 or

*u*+ 2 = 0, i.e.,

*u*= 0 or

*u*= - 2

Therefore, the zeroes of 4

*u*

^{2}+ 8

*u*are 0 and - 2.

Sum of zeroes = 0 + (-2) = -2 = -(8)/4 = -(Coefficient of

*u*)/Coefficient of

*u*

^{2}

Product of zeroes = 0 × (-2) = 0 = 0/4 = Constant term/Coefficient of

*u*

^{2}.

(v)

*t*

^{2}– 15

=

*t*

^{2 }- 0.

*t*- 15

= (

*t*- √15) (

*t*+ √15)

The value of

*t*

^{2}- 15 is zero when

*t*- √15 = 0 or

*t*+ √15 = 0, i.e., when

*t*= √15 or

*t*= -√15

Therefore, the zeroes of

*t*

^{2}- 15 are √15 and -√15.Sum of zeroes = √15 + -√15 = 0 = -0/1 = -(Coefficient of

*t*)/Coefficient of

*t*

^{2}

Product of zeroes = (√15) (-√15) = -15 = -15/1 = Constant term/Coefficient of

*u*

^{2}.

(vi) 3

*x*

^{2}–

*x*– 4

= (3

*x*- 4) (

*x*+ 1)

The value of 3

*x*

^{2}–

*x*– 4 is zero when 3

*x*- 4 = 0 and

*x*+ 1 = 0,i.e., when

*x*= 4/3 or

*x*= -1

Therefore, the zeroes of 3

*x*

^{2}–

*x*– 4 are 4/3 and -1.

Sum of zeroes = 4/3 + (-1) = 1/3 = -(-1)/3 = -(Coefficient of

*x*)/Coefficient of*x*^{2}
Product of zeroes = 4/3 × (-1) = -4/3 = Constant term/Coefficient of

2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i) 1/4 , -1*x*^{2}.2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(ii) √2 , 1/3

(iii) 0, √5

(iv) 1,1

(v) -1/4 ,1/4

(vi) 4,1

**Answer**

(i) 1/4 , -1

Let the polynomial be

*ax*

^{2}+

*bx*+

*c*, and its zeroes be Î± and ÃŸ

Î± + ÃŸ = 1/4 = -

*b*/

*a*

Î±ÃŸ = -1 = -4/4 =

*c*/

*a*

If

*a*= 4, then

*b*= -1,

*c*= -4

Therefore, the quadratic polynomial is 4

*x*

^{2}-

*x*-4.

(ii) √2 , 1/3

Let the polynomial be

*ax*

^{2}+

*bx*+

*c*, and its zeroes be Î± and ÃŸ

Î± + ÃŸ = √2 = 3√2/3 = -

*b*/

*a*

Î±ÃŸ = 1/3 =

*c*/

*a*

If

*a*= 3, then

*b*= -3√2,

*c*= 1

Therefore, the quadratic polynomial is 3

*x*

^{2}-3√2

*x*+1.

(iii) 0, √5

Let the polynomial be

*ax*

^{2}+

*bx*+

*c*, and its zeroes be Î± and ÃŸ

Î± + ÃŸ = 0 = 0/1 = -

*b*/

*a*

Î±ÃŸ = √5 = √5/1 =

*c*/

*a*

If

*a*= 1, then

*b*= 0,

*c*= √5

Therefore, the quadratic polynomial is

*x*

^{2}+ √5.

(iv) 1, 1

Let the polynomial be

*ax*

^{2}+

*bx*+

*c*, and its zeroes be Î± and ÃŸ

Î± + ÃŸ = 1 = 1/1 = -

*b*/

*a*

Î±ÃŸ = 1 = 1/1 =

*c*/

*a*

If

*a*= 1, then

*b*= -1,

*c*= 1

Therefore, the quadratic polynomial is

*x*

^{2}-

*x*+1.

(v) -1/4 ,1/4

Let the polynomial be

*ax*

^{2}+

*bx*+

*c*, and its zeroes be Î± and ÃŸ

Î± + ÃŸ = -1/4 = -

*b*/

*a*

Î±ÃŸ = 1/4 =

*c*/

*a*

If

*a*= 4, then

*b*= 1,

*c*= 1

Therefore, the quadratic polynomial is 4

*x*

^{2}+

*x*

(vi) 4,1

Let the polynomial be

*ax*

^{2}+

*bx*+

*c*, and its zeroes be Î± and ÃŸ

Î± + ÃŸ = 4 = 4/1 = -

*b*/

*a*

Î±ÃŸ = 1 = 1/1 =

*c*/

*a*

If

*a*= 1, then

*b*= -4,

*c*= 1

Therefore, the quadratic polynomial is

*x*

^{2}- 4

*x*

Page No: 36

**Exercise 2.3**

1. Divide the polynomial

*p*(

*x*) by the polynomial

*g*(

*x*) and find the quotient and remainder in each of the following:

^{}

**Answer**

(i)

*p*(

*x*) =

*x*

^{3}– 3

*x*

^{2}+ 5

*x*– 3,

*g*(

*x*) =

*x*

^{2}– 2

Quotient =

*x*-3 and remainder 7

*x*- 9

(ii)

Quotient = *p*(*x*) =*x*^{4}– 3*x*^{2}+ 4x + 5,*g*(*x*) =*x*^{2}+ 1 –*x**x*

^{2}+

*x*- 3 and remainder 8

(iii)

Quotient = -*p*(*x*) =*x*^{4}– 5*x*+ 6,*g*(*x*) = 2 –*x*^{2}*x*

^{2}-2 and remainder -5

*x*+10

^{}

second polynomial by the first polynomial:

**Answer**

(i)

*t*^{2}– 3, 2*t*^{4}+ 3*t*^{3}– 2*t*^{2}– 9*t*– 12*t*^{2}– 3 exactly divides 2*t*^{4}+ 3*t*^{3}– 2*t*^{2}– 9*t*– 12 leaving no remainder. Hence, it is a factor of 2*t*^{4}+ 3*t*^{3}– 2*t*^{2}– 9*t*– 12.*x*

^{2}+ 3

*x*+ 1, 3

*x*

^{4}+ 5

*x*

^{3}– 7

*x*

^{2}+ 2

*x*+ 2

*x*

^{2}+ 3

*x*+ 1 exactly divides 3

*x*

^{4}+ 5

*x*

^{3}– 7

*x*

^{2}+ 2

*x*+ 2 leaving no remainder. Hence, it is factor of 3

*x*

^{4}+ 5

*x*

^{3}– 7

*x*

^{2}+ 2

*x*+ 2.

*x*

^{3}– 3

*x*+ 1,

*x*

^{5}– 4

*x*

^{3}+

*x*

^{2}+ 3

*x*+ 1

*x*

^{3}– 3

*x*+ 1 didn't divides exactly

*x*

^{5}– 4

*x*

^{3}+

*x*

^{2}+ 3

*x*+ 1 and leaves 2 as remainder. Hence, it not a factor of

*x*

^{5}– 4

*x*

^{3}+

*x*

^{2}+ 3

*x*+ 1.

3. Obtain all other zeroes of 3

*x*^{4}+ 6*x*^{3}– 2*x*^{2}– 10*x*– 5, if two of its zeroes are √(5/3)**Answer**

*p*(

*x*) = 3

*x*

^{4}+ 6

*x*

^{3}– 2

*x*

^{2}– 10

*x*– 5

Since the two zeroes are √(5/3) and - √(5/3).

We factorize

*x*^{2}+ 2*x*+ 1
= (

*x*+ 1)^{2}
Therefore, its zero is given by

*x*+ 1 = 0*x*= -1

*x*+ 1)

^{2}, therefore, there will be 2 zeroes at

*x*= - 1.

Hence, the zeroes of the given polynomial are √(5/3) and - √(5/3), - 1 and - 1.

4. On dividing

*x*

^{3}- 3

*x*

^{2}+

*x*+ 2 by a polynomial

*g*(

*x*), the quotient and remainder were

*x*- 2 and

-2

*x*+ 4, respectively. Find

*g*(

*x*).

**Answer**

Here in the given question,

Dividend =

*x*^{3}- 3*x*^{2}+*x*+ 2
Quotient =

*x*- 2
Remainder = -2

*x*+ 4
Divisor =

*g*(*x*)
We know that,

Dividend = Quotient × Divisor + Remainder

⇒

⇒

⇒

∴

5.Give examples of polynomial

(i) deg

(ii) deg

(iii) deg

(ii) Let us assume the division of

Here,

Clearly, the degree of

Checking for division algorithm,

Thus, the division algorithm is satisfied.

(iii) Let us assume the division of

Here,

g(x) = x

Clearly, the degree of

Checking for division algorithm,

Thus, the division algorithm is satisfied.

Page No: 36

1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:

(i) 2

(ii)

(i)

Now for zeroes, putting the given value in x.

= (2×1/8) + 1/4 - 5/2 + 2

= 1/4 + 1/4 - 5/2 + 2

= 1/2 - 5/2 + 2 = 0

= (2×1) + 1 - 5 + 2

= 2 + 1 - 5 + 2 = 0

= (2 × -8) + 4 + 10 + 2

= -16 + 16 = 0

Thus, 1/2, 1 and -2 are the zeroes of the given polynomial.

Comparing the given polynomial with a

Also, Î±=1/2, Î²=1 and Î³=-2

Now,

-b/a = Î±+Î²+Î³

⇒ 1/2 = 1/2+ 1 - 2

⇒ 1/2 = 1/2

c/a = Î±Î²+Î²Î³+Î³Î±

⇒ -5/2 = (1/2 × 1) + (1 × -2) + (-2 × 1/2)

⇒ -5/2 = 1/2 - 2 - 1

⇒ -5/2 = -5/2

-d/a = Î±Î²Î³

⇒ -2/2 = (1/2 × 1 × -2)

⇒ -1 = 1

Thus, the relationship between zeroes and the coefficients are verified.

(ii)

Now for zeroes, putting the given value in x.

= 8 - 16 + 10 - 2

= 0

= 1 - 4 + 5 - 2

= 0

= 1 - 4 + 5 - 2

= 0

Thus, 2, 1 and 1 are the zeroes of the given polynomial.

Comparing the given polynomial with a

Also, Î±=2, Î²=1 and Î³=1

Now,

-b/a = Î±+Î²+Î³

⇒ 4/1 = 2+ 1 + 1

⇒ 4 = 4

c/a = Î±Î²+Î²Î³+Î³Î±

⇒ 5/1 = (2 × 1) + (1 × 1) + (1 × 2)

⇒ 5 = 2 + 1 + 2

⇒ 5 = 5

-d/a = Î±Î²Î³

⇒ 2/1 = (2 × 1 × 1)

⇒ 2 = 2

Thus, the relationship between zeroes and the coefficients are verified.

2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively.

Let the polynomial be

Then, Î± + Î² + Î³ = -(-2)/1 = 2 = -b/a

Î±Î² + Î²Î³ + Î³Î± = -7 = -7/1 = c/a

Î±Î²Î³ = -14 = -14/1 = -d/a

∴ a = 1, b = -2, c = -7 and d = 14

So, one cubic polynomial which satisfy the given conditions will be

3. If the zeroes of the polynomial

Since, (a - b), a, (a + b) are the zeroes of the polynomial

Therefore, sum of the zeroes = (a - b) + a + (a + b) = -(-3)/1 = 3

⇒ 3a = 3 ⇒ a =1

∴ Sum of the products of is zeroes taken two at a time = a(a - b) + a(a + b) + (a + b) (a - b) =1/1 = 1

a

⇒ 3a

Putting the value of a,

⇒ 3(1)

⇒ 3 - b

⇒ b

⇒ b = ±√2

Hence, a = 1 and b = ±√2

4. If two zeroes of the polynomial x

2+√3 and 2-√3 are two zeroes of the polynomial p(x) = x

Let x = 2±√3

So, x-2 = ±√3

On squaring, we get x

⇒ x

Now, dividing p(x) by x

∴ p(x) = x

= (x

= (x

= (x

= (x

∴ (x + 5) and (x - 7) are other factors of p(x).

∴ - 5 and 7 are other zeroes of the given polynomial.

*x*^{3}- 3*x*^{2}+*x*+ 2 = (*x*- 2) ×*g*(*x*) + (-2*x*+ 4)⇒*x*^{3}- 3*x*^{2}+*x*+ 2 - (-2*x*+ 4) = (*x*- 2) ×*g*(*x*)⇒

*x*^{3}- 3*x*^{2}+ 3*x*- 2 = (*x*- 2) ×*g*(*x*)⇒

*g*(*x*) = (*x*^{3}- 3*x*^{2}+ 3*x*- 2)/(*x*- 2)*g*(*x*) = (*x*^{2}-*x*+ 1)5.Give examples of polynomial

*p*(*x*),*g*(*x*),*q*(*x*) and*r*(*x*), which satisfy the division algorithm and(i) deg

*p*(*x*) = deg*q*(*x*)(ii) deg

*q*(*x*) = deg*r*(*x*)(iii) deg

*r*(*x*) = 0**Answer**
(i) Let us assume the division of 6

Here,

Degree of

Checking for division algorithm,

Or, 6

Hence, division algorithm is satisfied.

*x*^{2}+ 2*x*+ 2 by 2Here,

*p*(*x*) = 6*x*^{2}+ 2*x*+ 2*g*(*x*) = 2*q*(*x*) = 3*x*^{2}+*x*+ 1*r*(*x*) = 0Degree of

*p*(*x*) and*q*(*x*) is same i.e. 2.Checking for division algorithm,

*p*(*x*) =*g*(*x*) ×*q*(*x*) +*r*(*x*)Or, 6

*x*^{2}+ 2*x*+ 2 = 2*x*(3*x*^{2}+*x*+ 1)Hence, division algorithm is satisfied.

(ii) Let us assume the division of

*x*^{3}+*x*by*x*^{2},Here,

*p*(*x*) =*x*^{3}+*x**g*(*x*) =*x*^{2}*q*(*x*) =*x*and*r*(*x*) =*x*Clearly, the degree of

*q*(*x*) and*r*(*x*) is the same i.e., 1.Checking for division algorithm,

*p*(*x*) =*g*(*x*) ×*q*(*x*) +*r*(*x*)*x*^{3}+*x*= (*x*^{2}) ×*x*+*x**x*^{3}+*x*=*x*^{3}+*x*Thus, the division algorithm is satisfied.

(iii) Let us assume the division of

*x*^{3}+ 1 by*x*^{2}.Here,

*p*(*x*) =*x*^{3}+ 1g(x) = x

^{2}*q*(*x*) =*x*and*r*(*x*) = 1Clearly, the degree of

*r*(*x*) is 0.Checking for division algorithm,

*p*(*x*) =*g*(*x*) ×*q*(*x*) +*r*(*x*)*x*^{3}+ 1 = (*x*^{2}) ×*x*+ 1*x*^{3}+ 1 =*x*^{3}+ 1Thus, the division algorithm is satisfied.

Page No: 36

**Exercise 2.4 (Optional)**1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:

(i) 2

*x*^{3}+*x*^{2}- 5*x*+ 2; 1/2, 1, -2(ii)

*x*^{3}- 4*x*^{2}+ 5*x -*2; 2, 1, 1**Answer**(i)

*p*(*x*) = 2*x*^{3}+*x*^{2}- 5*x*+ 2Now for zeroes, putting the given value in x.

*p*(*1/2*) = 2(1/2)^{3}+*(1/2)*^{2}- 5(1/2)*+ 2*= (2×1/8) + 1/4 - 5/2 + 2

= 1/4 + 1/4 - 5/2 + 2

= 1/2 - 5/2 + 2 = 0

*p*(*1*) = 2(1)^{3}+*(1)*^{2}- 5(1)*+ 2*= (2×1) + 1 - 5 + 2

= 2 + 1 - 5 + 2 = 0

*p*(*-2*) = 2(-2)^{3}+ (-2)^{2}- 5(-2)*+ 2*= (2 × -8) + 4 + 10 + 2

= -16 + 16 = 0

Thus, 1/2, 1 and -2 are the zeroes of the given polynomial.

Comparing the given polynomial with a

*x*^{3}+ b*x*^{2}+ c*x*+ d, we get a=2, b=1, c=-5, d=2Also, Î±=1/2, Î²=1 and Î³=-2

Now,

-b/a = Î±+Î²+Î³

⇒ 1/2 = 1/2

⇒ -5/2 = (1/2 × 1) + (1 × -2) + (-2 × 1/2)

⇒ -5/2 = 1/2 - 2 - 1

⇒ -5/2 = -5/2

-d/a = Î±Î²Î³

⇒ -2/2 = (1/2 × 1 × -2)

⇒ -1 = 1

Thus, the relationship between zeroes and the coefficients are verified.

(ii)

*p*(*x*) =*x*^{3}- 4*x*^{2}+ 5*x -*2Now for zeroes, putting the given value in x.

*p*(*2*) = 2^{3}- 4*(2)*^{2}+ 5(2)*- 2*= 8 - 16 + 10 - 2

= 0

*p*(*1*) = 1^{3}- 4*(1)*^{2}+ 5(1)*- 2*= 1 - 4 + 5 - 2

= 0

*p*(*1*) = 1^{3}- 4*(1)*^{2}+ 5(1)*- 2*= 1 - 4 + 5 - 2

= 0

Thus, 2, 1 and 1 are the zeroes of the given polynomial.

Comparing the given polynomial with a

*x*^{3}+ b*x*^{2}+ c*x*+ d, we get a=1, b=-4, c=5, d=-2Also, Î±=2, Î²=1 and Î³=1

Now,

-b/a = Î±+Î²+Î³

⇒ 4/1 = 2

⇒ 5/1 = (2 × 1) + (1 × 1) + (1 × 2)

⇒ 5 = 2 + 1 + 2

⇒ 5 = 5

-d/a = Î±Î²Î³

⇒ 2/1 = (2 × 1 × 1)

⇒ 2 = 2

Thus, the relationship between zeroes and the coefficients are verified.

2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively.

**Answer**Let the polynomial be

*ax*^{3}+*bx*^{2 }+*cx + d*and the zeroes be Î±, Î² and Î³Then, Î± + Î² + Î³ = -(-2)/1 = 2 = -b/a

Î±Î² + Î²Î³ + Î³Î± = -7 = -7/1 = c/a

Î±Î²Î³ = -14 = -14/1 = -d/a

∴ a = 1, b = -2, c = -7 and d = 14

So, one cubic polynomial which satisfy the given conditions will be

*x*^{3}- 2*x*^{2 }-*7x + 14*3. If the zeroes of the polynomial

*x*^{3}– 3*x*^{2}+*x*+ 1 are a–b, a, a+b, find a and b.**Answer**Since, (a - b), a, (a + b) are the zeroes of the polynomial

*x*^{3}– 3*x*^{2}+*x*+ 1.Therefore, sum of the zeroes = (a - b) + a + (a + b) = -(-3)/1 = 3

⇒ 3a = 3 ⇒ a =1

∴ Sum of the products of is zeroes taken two at a time = a(a - b) + a(a + b) + (a + b) (a - b) =1/1 = 1

a

^{2}- ab + a^{2}+ ab + a^{2}- b^{2 }= 1⇒ 3a

^{2}- b^{2}=1Putting the value of a,

⇒ 3(1)

^{2}- b^{2}= 1⇒ 3 - b

^{2}= 1⇒ b

^{2}= 2⇒ b = ±√2

Hence, a = 1 and b = ±√2

4. If two zeroes of the polynomial x

^{4}– 6x^{3}– 26x^{2}+ 138x – 35 are 2±√3, find other zeroes.**Answer**2+√3 and 2-√3 are two zeroes of the polynomial p(x) = x

^{4}– 6x^{3}– 26x^{2}+ 138x – 35.Let x = 2±√3

So, x-2 = ±√3

On squaring, we get x

^{2}- 4x + 4 = 3,⇒ x

^{2}- 4x + 1= 0^{2}- 4x + 1^{4}- 6x^{3}- 26x^{2}+ 138x - 35= (x

^{2}- 4x + 1) (x^{2}- 2x - 35)= (x

^{2}- 4x + 1) (x^{2}- 7x + 5x - 35)= (x

^{2}- 4x + 1) [x(x - 7) + 5 (x - 7)]= (x

^{2}- 4x + 1) (x + 5) (x - 7)∴ (x + 5) and (x - 7) are other factors of p(x).

∴ - 5 and 7 are other zeroes of the given polynomial.

5. If the polynomial x

^{4}– 6x^{3}+ 16x^{2}– 25x + 10 is divided by another polynomial x^{2}– 2x + k, the remainder comes out to be x + a, find k and a.**Answer**

On dividing x

^{4}– 6x^{3}+ 16x^{2}– 25x + 10 by x^{2}– 2x + k
∴ Remainder = (2k - 9)x - (8 - k)k + 10

But the remainder is given as x+ a.

On comparing their coefficients,

2k - 9 = 1

⇒ k = 10

⇒ k = 5 and,

-(8-k)k + 10 = a

⇒ a = -(8 - 5)5 + 10 =- 15 + 10 = -5

Hence, k = 5 and a = -5

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thnx 4 ur effort.....its wonderful.....plz add question number 4 too for further look!

ReplyDeleteHi Obaidullah,

DeleteThanks for dropping by and helping us in making this platform more useful. We have added the required question. Your further suggestions are welcome.

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Study Rankers it will be very useful if u will add exemplar quetions

Deletethanks in advance

Thanks for your suggestion. We will look forward to it. Keep learning.

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can u please add ex 2.4 also

ReplyDeleteGood effort

ReplyDeleteStudy rankers this website is amazing but can you add maths of class 7also..

ReplyDeleteWaiting for the launch of mobile app

please + ex 2.4

ReplyDelete