Question 619186
For the quadratic equation
{{{ax^2+bx+c= 0}}}
The product of the zeros is c/a.<br>
For your expression, in this form, is:
{{{5z^2+13z+(-p)}}}
This makes your
a = 5
b = 13
c = -p<br>
You are told that the zeros of your expression are reciprocals. Zeros of a polynomial are the values for the variable that make the polynomial equal to zero. IOW, they are soluti8ons to:
{{{5z^2+13z+(-p) = 0}}}<br>
Inserting your "a" and "c" into c/a:
{{{(-p)/5}}}
Since your zeros are reciprocals are reciprocals and since the product of any reciprocals is always a 1, c/a must be 1:
{{{(-p)/5 = 1}}}<br>
Now we just solve this for p. Multiplying each side by 5:
{{{-p = 5}}}
Multiplying (or dividing) both sides by -1:
{{{p = -5}}}