Question

1)If alpha and beta are zeroes of quadratic polynomial x2 -(k+6x)+2(2k-1)
find k if alpha +beta=1/2alpha beta
2)If alpha and beta are zeroes of x2-6x+a, find the value of a if 3alpha+2beta=20

Answer 1

1.

$\textbf{Given:}$

$\text{ \alpha and \beta are zeros of x^2-(k+6)x+2(2k-1) }$

$\textbf{To find:}$

$\text{The value of k}$

$\textbf{Solution:}$

$\text{Since \alpha and \beta are zeros of x^2-(k+6)x+2(2k-1) , we have}$

$\alpha+\beta=\dfrac{-b}{a}=\dfrac{k+6}{1}=k+6$

$\alpha\beta=\dfrac{c}{a}=\dfrac{2(2k-1)}{1}=2(2k-1)$

$\text{But,}\;\bf\alpha+\beta=\dfrac{1}{2}(\alpha\beta)$

$k+6=\dfrac{1}{2}(2(2k-1))$

$k+6=2k-1$

$\implies\bf\,k=7$

$\therefore\textbf{The value of k is 7}$

2.

$\textbf{Given:}$

$\text{ \alpha and \beta are zeros of x^2-6x+a }$

$\textbf{To find:}$

$\text{The value of a}$

$\textbf{Solution:}$

$\text{Since \alpha and \beta are zeros of x^2-6x+a , we have}$

$\alpha+\beta=\dfrac{-b}{a}=\dfrac{6}{1}=6$

$\alpha\beta=\dfrac{c}{a}=\dfrac{a}{1}=a$

$\text{But,}\;\bf3\,\alpha+2\,\beta=20$

$\implies\,\alpha+2(\alpha+\,\beta)=20$

$\implies\,\alpha+2(6)=20$

$\implies\,\alpha=20-12$

$\implies\,\alpha=8$

$\alpha+\beta=6\implies\,\beta=-2$

$\text{Then,}$

$a=\alpha\beta=(8)(-2)=-16$

$\therefore\textbf{The value of a is -16}$

Find more:

The equation whose roots are smaller by 1 than those of 2x2 – 5x + 6 = 0 is 

a) 2x^2– 9x + 13 = 0

b) 2x^2– x + 3 = 0 

c) 2x^2+ 9x + 13 = 0

d) 2x^2 + x + 3 = 0​

https://brainly.in/question/16821146

If alpha,beta are the roots of equation x^2-5x+6=0 and alpha > beta then the equation with the roots (alpha+beta)and (alpha-beta) is

https://brainly.in/question/5731128

Answer 2

Answer:

1 ans

Step-by-step explanation:

alpha + beta=-b/a

=k+6

alpha×beta=c/a

2(2k-1)/1

=4k-2

x^2-(k+6)x= 2(2k-1)

k+6=2(2k-1)/2

k+6=2k-1

2k-k=6+1

k=7