The rational root theorem states that the rational roots of a polynomial can only be in the form p/q, where p divides the constant term, and q divides the leading term.
In your case, both the leading term 5 and the constant term 11 are primes, so their only divisors are 1 and themselves.
So, the only feasible solutions are

For the record, in this case, none of the feasible solutions are actually a root of the polynomial.
Answer: <u>4 pounds</u> of brand X sugar
====================================================
Reason:
n = number of pounds of brand X sugar
5n = cost of buying those n pounds, at $5 per pound
Brand Y costs $2 per pound, and you buy 8 lbs of it, so that's another 2*8 = 16 dollars.
5n+16 = total cost of brand X and brand Y combined
n+8 = total amount of sugar bought, in pounds
3(n+8) = total cost because we buy n+8 pounds at $3 per pound
The 5n+16 and 3(n+8) represent the same total cost.
Set them equal to each other. Solve for n.
5n+16 = 3(n+8)
5n+16 = 3n+24
5n-3n = 24-16
2n = 8
n = 8/2
n = 4 pounds of brand X sugar are needed
-------------
Check:
n = 4
5n = 5*4 = 20 dollars spent on brand X alone
16 dollars spent on brand Y mentioned earlier
20+16 = 36 dollars spent total
n+8 = 4+8 = 12 pounds of both types of sugar brands combined
3*12 = 36 dollars spent on both types of sugar brands
The answer is confirmed.
--------------
Another way to verify:
5n+16 = 3(n+8)
5*4+16 = 3(4+8)
20+16 = 3(12)
36 = 36
Answer:
<u>c. no solution</u>
Step-by-step explanation:
This system of equations has no solution because when graphed, the lines are parallel to each other. Meaning they will never intersect. Since they will never intersect, there are no solutions.
Answer:
A
Step-by-step explanation:
Given
x² - x = 30
To complete the square
add ( half the coefficient of the x- term )² to both sides
x² + 2(-
)x +
= 30 + 
(x -
)² =
Take the square root of both sides
x -
= ±
= ± 
Add
to both sides
x =
±
, thus
x =
-
= - 5
x =
+
= 6
Is there any more information?