Question

if alpha and beta are the zeros of polynomial p(x)=6x^2-5x+k such that alpha - beta = 1/6 find the value of k

Answer 1

a=alpha
b= beta


so...

a+b = 5/6
and
a-b =1/6

so...
a+b=5/6
a-b =1/6
_______
2a. =1
a=1/2

p(x)=6x^2-5x+k
p(a)=6(1/2)^2-5(1/2)+k
0 =6(1/4)-5/2+k
0 =(3/2)-(5/2)+k
0=-(2/2)+k
k-1=0
so. k=1

Answer 2

Answer: The value of k is 1.

Step-by-step explanation:

Since we have given that

p(x) is given by

$6x^2-5x+k$

Here,

$\alpha -\beta =\dfrac{1}{6}--------------(1)$

As we know that

Sum of zeroes would be

$\alpha +\beta =\dfrac{5}{6}-------(2)$

From eq(1), and (2), we get that

$\beta +\dfrac{1}{6}+\beta =\dfrac{5}{6}\\\\2\beta =\dfrac{5}{6}-\dfrac{1}{6}\\\\2\beta =\dfrac{4}{6}\\\\\beta =\dfrac{2}{6}=\dfrac{1}{3}$

$\alpha=\dfrac{1}{6}+\beta =\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{1+2}{6}=\dfrac{3}{6}=\dfrac{1}{2}$

And

product of zeroes would be

$\alpha \beta =\dfrac{1}{2}\times \dfrac{1}{3}=\dfrac{1}{6}$

Hence,  the value of k is 1.