Question

please solve with full steps (only question 6)

Answer 1

Answer:

$\large \boxed{\sf \ \ k=1 \ \ }$

Step-by-step explanation:

Hello,

First of all, let's notice that even if we do not know the zeros of P(x) we can say that

       $(1) \ \alpha + \beta =\dfrac{5}{6} \\ \\ (2) \ \alpha * \beta =\dfrac{k}{6}$

But, why !?

As they are the zeros of P(x), we can write:

$P(x)=6(x^2-\boxed{\dfrac{5}{6}}x+\boxed{\dfrac{k}{6}})=6(x-\alpha)(x-\beta)=6(x^2-\boxed{(\alpha +\beta)}x+ \boxed{\alpha *\beta} )$

And then we can identify the coefficients of the like terms to find the equations (1) and (2).

Now, we have one more equation which is:

       $(3) \ \alpha -\beta =\dfrac{1}{6}$

(1)+(3) gives:

   $\alpha + \beta +\alpha -\beta =\dfrac{5}{6}+\dfrac{1}{6}=\dfrac{6}{6}=1 \\ \\<=> 2\alpha =1 \ \text{ divide by 2 } \\ \\ <=> \alpha =\dfrac{1}{2}$

And we replace in (3) to get the value of the second zero.

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

And, finally, from (2), it comes:

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

Hope this helps.

Do not hesitate if you need further explanation.

Thank you