Question

solve these all questions for 50 points!!!!
(small sums can be texted and long sums can be kept by adding a photo)....

Answer 1

14)
given polynomial,
p(x)=2x^2+kx-15
given that 3 is one of the zero of the given polynomial,
then,p(3)=0
2(3)^2+3k-15=0
18+3k-15=0
k=-1
Now the polynomial becomes,
2x^2-x-15=0
2x^2-6x+5x-15=0
(2x+5)(x-3)=0
Hence other zero =-5/2







15)
polynomial,ax^2+bx+c=0
let alpha and beta be the zeroes.
as per given condition we get,
alpha=2beta..............................1
sum of the zeroes =-b/a
alpha+beta=-b/a
3beta=-b/a. (by eq1 we get)
beta=-b/3a..................................2
product of the zeroes=c/a
alpha×beta=c/a
2beta^2=c/a
2(-b/3a)^2=c/a
2b^2/9a^2=c/a
2b^2/9a=c
2b^2=9ac
Hence proved.




16)
Given polynomial,
x^2+a=0
x^2+0x+a=0
let alpha and beta be the zeroes.
alpha =-3
sum of the zeroes =0
alpha +beta=0
-3+beta=0
beta=3.
product of the zeroes =a
alpha×beta=a
-3×3=a
a=-9
Therefore value of a=-9 and other zero is equals to 3.






17)
Given polynomial,
p(x)=x^2-5x+6
sum of the zeroes=alpha+beta=5
product of the zeroes =alpha×beta=6
now,
alpha+beta-3alphabeta=5-3(6)=5-18=-13
Hence value of,
a+b-3ab=-13






18)
Given polynomials,
p(x)=x^2+px+q ...............................1
and f(x)=2x^2-5x-3
f(x)=0
2x^2-5x-3=0
2x^2-6x+x-3=0
(2x+1)(x-3)=0
x=-1/2. or. x=3
let alpha and beta be the zeroes of p(x)
As per given condition we get
alpha =2(-1/2)=-1
beta=2(3)=6
we know that quadratic polynomial is in the form of,
=k{x^2-(alpha+beta)x+alphabeta}
=k{x^2-(-1+6)x+(-1)(6)}
=k{x^2-5x-6}
=x^2-5x-6........................................2
On comparing eq1 and eq2 we get,
value of p=-5 and q=-6





19)
given polynomial,
f(x)=x^2-5x+k
and a-B=1...............................1
sum of the zeroes =5
a+B=5.............................2
By eq1 and eq2 we get,
a=3. and B=2
product of the zeroes =k
a×B=k
k=6
Hence value of k =6







20)
Given polynomial,
f(x)=x^2-8x+k
let alpha and beta be the zeroes
alpha=a. and beta=B
as per given condition we get,
a^2+B^2=40
(a+B)^2-2aB=40
we know that,a+B=8 and aB=k
now we get,
(8)^2-2k=40
64-2k=40
-2k=-24
k=12
Hence value of k = 12





21)
Given polynomial
f(x)=3x^2-4x+1
By the method of factorization we get,
zeroes =1/3 and 1
a=1/3. and B=1
a^2/B=(1/3)^2/1=1/9
B^2/a=(1)^/1/3=3
we know that quadratic polynomial is in the form of,
=k{x^2-(sum of the zeroes)x + product of the zeroes}
=k{x^2-(a^2/B+B^2/a)x+a^2/B×B^2/a}
=k{x^2-(1/9+3)x+1/9×3}
=k{x^2-(28/9)x+1/3}
taking k=9 we get,
=9{x^2-28/9x+1/3}
=9x^2-28x+3
Hence 9x^2-28x+3 is the required polynomial.




23)
x^2-6x+5 is the required polynomial.





24)
Here p(x)=x^3-3x^2-x+3
g(x)=x-3
when p(x) is divided by g(x) we get,
q(x)=x^2-1
and r(x)=0 as it is a factor.
Dividend=Divisor×Quotient + remainder
Dividend=p(x)=x^3-3x^2-x+3
=Divisor×Quotient + remainder
=g(x)×q(x)+r(x)
=(x-3)(x^2-1)+0
=x^3-3x^2-x+3
=p(x)
=Dividend
Hence division algorithm verified.

Answer 2

______✨ HEY MATE ✨______

==========
Answer :-
==========

14) given polynomial,
p(x)=2x^2+kx-15
given that 3 is one of the zero of the given polynomial,
then,p(3)=0
2(3)^2+3k-15=0
18+3k-15=0
k=-1
Now the polynomial becomes,
2x^2-x-15=0
2x^2-6x+5x-15=0
(2x+5)(x-3)=0
Hence other zero =-5/2

15) polynomial,ax^2+bx+c=0
let alpha and beta be the zeroes.
as per given condition we get,
alpha=2beta..............................1
sum of the zeroes =-b/a
alpha+beta=-b/a
3beta=-b/a. (by eq1 we get)
beta=-b/3a..................................2
product of the zeroes=c/a
alpha×beta=c/a
2beta^2=c/a
2(-b/3a)^2=c/a
2b^2/9a^2=c/a
2b^2/9a=c
2b^2=9ac
Hence proved.

16)Given polynomial,
x^2+a=0
x^2+0x+a=0
let alpha and beta be the zeroes.
alpha =-3
sum of the zeroes =0
alpha +beta=0
-3+beta=0
beta=3.
product of the zeroes =a
alpha×beta=a
-3×3=a
a=-9
Therefore value of a=-9 and other zero is equals to 3.

17) Given polynomial,
p(x)=x^2-5x+6
sum of the zeroes=alpha+beta=5
product of the zeroes =alpha×beta=6
now,
alpha+beta-3alphabeta=5-3(6)=5-18=-13
Hence value of,
a+b-3ab=-13





Secondary School

Math

50+25 pts

Solve these all questions for 50 points!!!!
(small sums can be texted and long sums can be kept by adding a photo)....



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Comments (5) 

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 by Sindhusharavuri08.04.2017



it is difficult to solve at one place .sry



atleast solve any 5



plss



solve 14to 18 sums



then.

Answers



THE BRAINLIEST ANSWER!



Prajapatyk01 

Ace

14)
given polynomial,
p(x)=2x^2+kx-15
given that 3 is one of the zero of the given polynomial,
then,p(3)=0
2(3)^2+3k-15=0
18+3k-15=0
k=-1
Now the polynomial becomes,
2x^2-x-15=0
2x^2-6x+5x-15=0
(2x+5)(x-3)=0
Hence other zero =-5/2







15)
polynomial,ax^2+bx+c=0
let alpha and beta be the zeroes.
as per given condition we get,
alpha=2beta..............................1
sum of the zeroes =-b/a
alpha+beta=-b/a
3beta=-b/a. (by eq1 we get)
beta=-b/3a..................................2
product of the zeroes=c/a
alpha×beta=c/a
2beta^2=c/a
2(-b/3a)^2=c/a
2b^2/9a^2=c/a
2b^2/9a=c
2b^2=9ac
Hence proved.




16)
Given polynomial,
x^2+a=0
x^2+0x+a=0
let alpha and beta be the zeroes.
alpha =-3
sum of the zeroes =0
alpha +beta=0
-3+beta=0
beta=3.
product of the zeroes =a
alpha×beta=a
-3×3=a
a=-9
Therefore value of a=-9 and other zero is equals to 3.






17)
Given polynomial,
p(x)=x^2-5x+6
sum of the zeroes=alpha+beta=5
product of the zeroes =alpha×beta=6
now,
alpha+beta-3alphabeta=5-3(6)=5-18=-13
Hence value of,
a+b-3ab=-13






18) Given polynomials,
p(x)=x^2+px+q ...............................1
and f(x)=2x^2-5x-3
f(x)=0
2x^2-5x-3=0
2x^2-6x+x-3=0
(2x+1)(x-3)=0
x=-1/2. or. x=3
let alpha and beta be the zeroes of p(x)
As per given condition we get 
alpha =2(-1/2)=-1
beta=2(3)=6
we know that quadratic polynomial is in the form of,
=k{x^2-(alpha+beta)x+alphabeta}
=k{x^2-(-1+6)x+(-1)(6)}
=k{x^2-5x-6}
=x^2-5x-6........................................2
On comparing eq1 and eq2 we get,
value of p=-5 and q=-6

19)
given polynomial,
f(x)=x^2-5x+k
and a-B=1...............................1
sum of the zeroes =5
a+B=5.............................2
By eq1 and eq2 we get,
a=3. and B=2
product of the zeroes =k
a×B=k
k=6
Hence value of k =6


20)
Given polynomial,
f(x)=x^2-8x+k
let alpha and beta be the zeroes
alpha=a. and beta=B
as per given condition we get,
a^2+B^2=40
(a+B)^2-2aB=40
we know that,a+B=8 and aB=k
now we get,
(8)^2-2k=40
64-2k=40
-2k=-24
k=12
Hence value of k = 12

21) Given polynomial
f(x)=3x^2-4x+1
By the method of factorization we get,
zeroes =1/3 and 1
a=1/3. and B=1
a^2/B=(1/3)^2/1=1/9
B^2/a=(1)^/1/3=3
we know that quadratic polynomial is in the form of,
=k{x^2-(sum of the zeroes)x + product of the zeroes}
=k{x^2-(a^2/B+B^2/a)x+a^2/B×B^2/a}
=k{x^2-(1/9+3)x+1/9×3}
=k{x^2-(28/9)x+1/3}
taking k=9 we get,
=9{x^2-28/9x+1/3}
=9x^2-28x+3
Hence 9x^2-28x+3 is the required polynomial.




23) x^2-6x+5 is the required polynomial.

24) Here p(x)=x^3-3x^2-x+3
g(x)=x-3
when p(x) is divided by g(x) we get,
q(x)=x^2-1
and r(x)=0 as it is a factor.
Dividend=Divisor×Quotient + remainder
Dividend=p(x)=x^3-3x^2-x+3
=Divisor×Quotient + remainder
=g(x)×q(x)+r(x)
=(x-3)(x^2-1)+0
=x^3-3x^2-x+3
=p(x)
=Dividend
Hence division algorithm verified.

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