Question

if a and b are zeroes of the polynomial p(x)= 6x^2-5x+k such that a-b=1/6, find the value of k

Answer 1

SOLUTION

Given,

p(x)=6x^2-5x+k,such that alpha-beta=1/6

$\alpha + \beta = - \frac{b}{a} = \frac{5}{6} \\ \\ \alpha \beta = \frac{c}{a} = \frac{k}{6} \\ \\ \alpha - \beta = \frac{1}{6} \\ \\ = > {( \alpha - \beta )}^{2} = ( \frac{1}{6} ) {}^{2} \\ \\ = > ( \alpha + \beta ) {}^{2} - 4 \alpha \beta = ( \frac{1}{6} ) {}^{2} \\ \\ = > ( \frac{5}{6} ) {}^{2} - 4( \frac{k}{6} ) = \frac{1}{36} \\ \\ = > \frac{25}{36} - \frac{2k}{3} = \frac{1}{36} \\ \\ = > \frac{25 - 24k}{36} = \frac{1}{36} \\ \\ = > 25 - 24k = 1 \\ \\ = > 25 - 1 = 24k \\ \\ = > 24 = 24k \\ \\ = > k = 1$

Hope it helps ☺️

Answer 2

Answer:

Step-by-step explanation: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