Asked/Added Questions From NCERT:


Question 1. Asked on :19 April 2019:08:23:46 AM

 Check whether the first polynomials is a factors of the second polynomials by dividing the second polynomials by the first polynomials.

-Added by Prince Rawat Mathematics » POLYNOMIALS


Question 2. Asked on :27 January 2019:11:08:39 AM

 (4-k)x2+(2k+4)x+(8k+1)=0.

k=? 

एक पुर्ण वर्ग है ?

-Added by SURAJ KUMAR Mathematics » POLYNOMIALS

Answer:

Himanshi Verma

 The given quadratic will be a perfect square if it has two real and equal roots. To have two real and equal roots the discriminant must be zero.

i.e., b²-4ac=0

or, (2k+4)²-4(4-k)(8k+1)=0

or, 4k²+16k+16-4(32k-8k²+4-k)=0

o², 4k²+16k+16-128k+32k²-16+4k=0

or, 36k²-108k=0

or, 36k(k-3)=0

either 36k=0

or, k=0

or, k-3=0

or, k=3

∴, k=0,3 Ans.

-Answered by Himanshi Verma On 02 February 2019:01:43:20 PM


Question 3. Asked on :27 January 2019:10:59:13 AM

 abx2+(b2ac)x-bc=0.सरल करे|


-Added by SURAJ KUMAR Mathematics » POLYNOMIALS

Answer:

Himanshi Verma

 Consider, abx2 +(b2-ac)x-bc = 0

⇒ abx2 + b2x – acx – bc = 0
⇒ bx (ax + b) – c(ax + b) = 0
⇒ (ax + b)(bx – c) = 0

-Answered by Himanshi Verma On 02 February 2019:01:44:13 PM


Question 4. Asked on :27 January 2019:10:54:59 AM

 ax2+4ax+(a2-b2)=0

-Added by SURAJ KUMAR Mathematics » POLYNOMIALS


Question 5. Asked on :27 January 2019:10:51:05 AM

1/a+b+x=1/a+1/b+1/x,a+b not equal 0.सरल करे |  

-Added by SURAJ KUMAR Mathematics » POLYNOMIALS

Answer:

Himanshi Verma

ON solving the above equation we have a quadratic equation  

(a+b)x^2+(a+b)^2x+ab(a+b)

on solving the above equation by formula -b+root over(b^2-4ac) and -b-root over(b^2-4ac)

we got -a and -b as roots.

According to me this is much easier then others when compared.

 

-Answered by Himanshi Verma On 10 March 2019:04:49:29 PM


Question 6. Asked on :27 January 2019:10:46:14 AM

 

-Added by SURAJ KUMAR Mathematics » POLYNOMIALS


Question 7. Asked on :12 January 2019:10:49:51 AM

 Factorise:

x3-23x2+142x-120

-Added by Akki chauhan Mathematics » Polynomials

Answer:

Himanshi Verma

  x-23x2+142x-120 

=x3-x2-22x2+22x+120-120

=x2(x-1)-22x(x-1)+120(x-1)

=(x-1)(x2-22x+120)

=(x-1)(x2-12x-10x+120)

=(x-1)[x(x-12)-10(x-12)]

=(x-1)(x-10)(x-12)

-Answered by Himanshi Verma On 16 January 2019:05:57:48 PM


Question 8. Asked on :12 January 2019:10:45:37 AM

 If the polynomials ax3+3x2-13 and 2x3-5x+a , are divided by x+2 if the remainder in each case is the same, then find the value of 'a'

-Added by Akki chauhan Mathematics » Polynomials

Answer:

Himanshi Verma

  When p(x) is divided by (x-2),


By remainder theorem,


Remainder= p(2)


p(2)=a*2^3 + 3*2^2 - 13


        =a*8+3*4-13


       =8a+12-13 = 8a-1



p(2) = 2*2^3-5*2+a


      =2*8-10+a


     =16-10+a = 6+a


Since the remainders of ax^3+3x^2-13 and 2x^3-5x+a are same when divided by x-2,


8a-1 = 6+a


8a-6+a=1


8a+8=1-6=-5


8a=-5-8 =-13


a=-13/8

-Answered by Himanshi Verma On 20 January 2019:05:29:25 PM


Question 9. Asked on :12 January 2019:10:38:50 AM

If a = 9-4√5, then find the value of

a21a2 

-Added by Akki chauhan Mathematics » Polynomials

Answer:

Himanshi Verma

 (a+1/a)2=a2+1/a2+2

a2+1/a2=(a+1/a)2-2→ I

Now,

a=9-4root5

1/a=1/9-4root5

=(1/9-4root5)×9+4root5/9+4root5

=(9+4root5)

a+1/a=9-4root5+9+4root5

=81

Now,

a2+1/a2=(a+1/a)2-2 →(from I)

=(81)2-2

=6561-2

=6559


-Answered by Himanshi Verma On 22 January 2019:05:31:51 PM


Question 10. Asked on :11 January 2019:10:04:29 AM

 What is Degree of the Polynomials?

-Added by Prince Rawat Mathematics » POLYNOMIALS

Answer:

Prince Rawat

 Highest power of x in the algebraic expression is called the degree of the polynomials.

-Answered by Prince Rawat On 11 January 2019:10:08:00 AM


 

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