how to calculate roots of given equation below? Without solving equation $2x^2 + 9x + 9 = 0$, show that one of the root of the equation is twice the other.
 A: Suppose you have a quadratic equation $q(x)=0$ with one root twice the other $$q(x)=x^2+bx+c=(x-a)(x-2a)=x^2-3ax+2a^2$$ whence $b=-3a\dots(1)$ and $c=2a^2\dots (2)$.
To find a condition for $b$ and $c$ without knowing what $a$ is, you need to eliminate $a$ from equations $(1)$ and $(2)$. I'll leave it at that for the moment because your question shows no work or ideas - but note you will have to show the condition is sufficient as well as necessary. This is easy once you know what it is.
A: let the roots be $\alpha$ and $k$ . 
$$\alpha + k = -9/2$$
$$\alpha \times k =  9/2$$
$$\alpha \times ( 1 + k ) = 9/2$$  and  $$k\alpha^2  = 9/2( 1+k )^2/k = ( 81/4 )\times ( 2/9 )$$
$$( 1+2k+2k^2)/k = 9/2$$
$$2k^2 - 5k + 2 =0$$
$$(2k -1)(k-2)=0$$
$k=1/2$   or  $k=2.$
A: Mark's answer explains that every quardratic equation s.t one of the roots is twice the other is of a general expression. So we can just show the result by comparing the genral expression with the equation given in the question.
My answer is similar to this.  

If we are not allowed to solve the given equation then we we can show this by reversing the situation, that is first conjecture that the equation has two roots s.t one is twice the other.
Let $\alpha$ and $\beta$ be the roots of the given equation s.t $\alpha=2\beta$.  $$if\ \ \ \ \ \ \ (x-2\beta)(x-\beta)=0$$
$$then\ \ \ \ x^2-\beta x-2\beta x+2\beta^2=0$$ $$then\ \ \ \ x^2-3\beta x+4\beta^2=0 \tag{1}$$
Different values of $\beta$ will give different expressions so we can put different values of $\beta$ by hit and trial to see whether our conjecture is true or not.
Let's put $\beta={-3/2}$ then equation $1$ reduces to :$$ 2x^2 + 9x + 9 = 0$$
Which is the required equation hence our supposition is true .
