How do I calculate these questions relating to roots of quadratic equation?

[eureka_beyond]

Homework Statement

These two questions are very similar:
1) Let c be a constant. If a and b are the roots of the equation x^2 + 2x - c = 0
then 2b-a^2 = ?

2) Let k be a constant. If a and b are the roots of the equation x^2 - 3x + k = 0
Then a^2 + 3b = ?

Homework Equations

The answer to these 2 questions are -4-c and 9-k.

The Attempt at a Solution

For the first one, I was thinking about using the sum of roots and product of roots, I know that a+ b is -2 and ab must be c, but I don't know how to convert 2b-a^2 into a new equation that only consists of a+b and ab . I used another approach, which is to turn the equation into 2 separate ones by knowing the two roots given, a^2 + 2a - c = 0 and b^2 + 2b - c = 0. Then after some change of subject, I'll know that a^2 = c-2a and 2b = c-b^2 . But if I put in these two together, 2b-a^2 will equal to c-(b^2)-c+2a, which will finally give 2a-b^2, which doesn't really solve the problem...
I thought of another way, which is to let those unknowns to be a random number. Since this is a multiple choice question, I can get the answer by try and error. However, I was thinking if there's a better way.
Thanks

 

[Dick]

Your first approach was right. a+b=(-2). So b=(-2)-a. 2b-a^2=2(-2-a)-a^2=(-4)-(2a+a^2). Now finish up.

 

[Architeuthis Dux]

I would suggest using the quadratic formula to solve these equations. The quadratic formula is a general formula for finding the roots of a quadratic equation, which can be written as ax^2 + bx + c = 0. The formula is:

x = (-b ± √(b^2 - 4ac)) / 2a

In the first equation, x^2 + 2x - c = 0, we can see that a = 1, b = 2, and c = -c. Plugging these values into the quadratic formula, we get:

x = (-2 ± √(2^2 - 4(1)(-c))) / 2(1)
x = (-2 ± √(4 + 4c)) / 2
x = (-2 ± 2√(1 + c)) / 2
x = (-1 ± √(1 + c))

Since a and b are the roots of this equation, we can substitute them into the formula to get:

a = (-1 + √(1 + c))
b = (-1 - √(1 + c))

Now, we can plug these values into the expressions we are trying to solve:

2b - a^2 = 2(-1 - √(1 + c)) - (-1 + √(1 + c))^2
= -2 - 2√(1 + c) - (1 + c + 2√(1 + c))
= -2 - 1 - c - 2√(1 + c) - 2√(1 + c)
= -3 - c - 4√(1 + c)

Using the same approach for the second equation, x^2 - 3x + k = 0, we get:

a = (3 + √(9 - 4k)) / 2
b = (3 - √(9 - 4k)) / 2

Plugging these values into the expression, we get:

a^2 + 3b = (3 + √(9 - 4k))^2 / 4 + 3((3 - √(9 - 4k)) / 2)
= (9 + 6√(9 - 4k) + (9 - 4k