Question

plz answer these questions...

question second and third. ​

Answer 1

Step-by-step explanation:

2) Sum of zeros = p+q = -{-(k+6)}/1= k+6

and , product of zeros = pq = 2(2k-1)/1= 4k-2.

From the question ,

2(p+q)= pq

→2(k+6)= 4k-2

→2k+12= 4k-2

→-2k= -14

→k= 7. ...

Answer 2

Step-by-step explanation:

2) Sum of zeros = p+q = -{-(k+6)}/1= k+6

and , product of zeros = pq = 2(2k-1)/1= 4k-2.

From the question ,

2(p+q)= pq

→2(k+6)= 4k-2

→2k+12= 4k-2

→-2k= -14

→k= 7. ...

3) Given, p(x) = 2x3+x2-5x+2

And zeroes for p(x) are = 1/2, 1, -2

∴ p(1/2) = 2(1/2)3+(1/2)2-5(1/2)+2 = (1/4)+(1/4)-(5/2)+2 = 0

p(1) = 2(1)3+(1)2-5(1)+2 = 0

p(-2) = 2(-2)3+(-2)2-5(-2)+2 = 0

Hence, proved 1/2, 1, -2 are the zeroes of 2x3+x2-5x+2.

Now, comparing the given polynomial with general expression, we get;

∴ ax3+bx2+cx+d = 2x3+x2-5x+2

a=2, b=1, c= -5 and d = 2

As we know, if α, β, γ are the zeroes of the cubic polynomial ax3+bx2+cx+d , then;

α +β+γ = –b/a

αβ+βγ+γα = c/a

α βγ = – d/a.

Therefore, putting the values of zeroes of the polynomial,

α+β+γ = ½+1+(-2) = -1/2 = –b/a

αβ+βγ+γα = (1/2×1)+(1 ×-2)+(-2×1/2) = -5/2 = c/a

α β γ = ½×1×(-2) = -2/2 = -d/a

Hence, the relationship between the zeroes and the coefficients are satisfied.