Easy polynomials question? Please do this without using the quadratic formula.
If $\alpha$ and $\beta$ are zeroes of the polynomial $x^2 -6x + a$ then find the value of "$a$" if $3\times \alpha + 2\times \beta = 20$
Thank you for the help
There is also a second question of this sort, i dont get that either. Would help if both were answered.
if $\alpha$ and $\beta$ are zeroes of the polynomial $x^2 -5x + k$ such that alpha- beta = 1. Find K 
 A: First of all don't confuse $a$ with $\alpha$ !(It is better to substitute A or m for $a$)
We know that the sum of the roots of a quadratic equation $ax^2+bx+c$ is $x_1+x_2=-\frac{b}{a}$
So: $\alpha +\beta =6$
On the other hand:$3\times \alpha + 2\times \beta =20$
Solving this two equations two variables yields:$\beta = -2$ and $\alpha=8$
We also know that the product of the roots of a quadratic equation $ax^2+bx+c$ is  : $x_1 \times x_2=\frac{c}{a}$
so $$\alpha \times \beta = -16 = a  $$  
Second question:
Again:
We know that the sum of the roots of a quadratic equation $ax^2+bx+c$ is $x_1+x_2=-\frac{b}{a}$
So: $\alpha +\beta =5$
On the other hand:$\alpha - \beta =1$
Solving this two equations two variables yields:$\beta = 2$ and $\alpha=3$
We also know that the product of the roots of a quadratic equation $ax^2+bx+c$ is  : $x_1 \times x_2=\frac{c}{a}$
so $$\alpha \times \beta = 6 = k  $$  
A: In the second question, we have $\alpha+\beta=5$ and $\alpha-\beta=1$. Note the general identity
$$(\alpha+\beta)^2-(\alpha-\beta)^2=4\alpha\beta,\tag{$1$}$$
which can be easily verified by expanding the squares. Putting $\alpha+\beta=5$ and $\alpha-\beta=1$ we get $4\alpha\beta=24$, so $\alpha\beta=6$.
Remark: The solution procedure by PooyaM is better, since it works in both questions. However, the identity $(1)$ is sometimes useful, so I thought I would mention it.
