Question

pls answer all I will mark brainly and follow ​

Answer 1

Step-by-step explanation:

2x^2 -5x+8

Here,

A= 2

B= -5

C= 8

sum of zeroes = -b/a

Alpha +beta = -5/2.…..(1)

And,

product of zeroes = c/a

Alpha ×beta = 8/2...(2)

To find ,

(Alpha + beta )^2- 2× alpha ×beta

(5/2)^2 -2×4

25/4-8

25-32/4

-7/4

2.since , one of the zeros of the polynomial

3x^2 +(2k+7)x-4 is negative of the other .

Let alpha and beta be the zeroes of 3x^2 +(2x+7)x-4

so,

Alpha = -beta .so , a+b=0

-(2k+7)x/3=0

so, k=-7/2

so ,3x^2 -4= 0

so, x= 2√3/3

3. sum of zeroes of the given polynomial=product of its zeroes

-b/a =c/a

-2/k =3k/k

k= -2/3

4.f(x)= (k^2 +4)x^2 +13x +4k

Now ,let alpha and beta be the roots ,then according to question ,

a= 1/b

= AB =1

now ,we know that Ab = 4k/k^2 +4

1= 4k/k^2+4

k^2 -4k+4=0

(k-2)^2 = 0

k-2= 0

k= 2

5.since , alpha and beta are the zeroes of the quadratic polynomial f(x)= x^2-p(x+1)-c

Then ,

x^2 -p(x+1)-c

x^2 -px -p-c

alpha +beta = -b/a

= -(-p)/1

= p

Ab= c/a

-p-c/1

-p-c

we have to prove that

(a+1) (B+1)

taking a and b because of incomplete statement

(a+1) (B+1)= 1-c

(a+1)B + (a+1)(1)= 1-c

AB + B+A +1 = 1-c

AB +(a+B )+1 =1-c

substituting a+B = p and aB = -p-c we get ,

-p-c+p+1= 1-c

1-c= 1-c

hence ,it is shown that (a+1) (B+1)= 1-c

I hope it will help you