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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

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    At least get the variables corrected. Alpha is not 'a' and beta is undefined.2012-09-01

2 Answers 2

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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 $

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    @EmmadKareem: As It is mentioned in the answer:We know that the **sum** of the root of a quadratic equation $ax^2+bx+c$ is $x_1+x_2=-\frac{b}{a}$ And for proving this, you should use the general quadratic formula(If you meant this)2012-09-01
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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.