The roots of quadratic equation $3x^2 - 5x - k = 0$ are $m/3$ and $m + 3$. Find the values of $m$ and $k$ Can someone please help me solve this equation. Please put in your working as well so I can see how to do this because this equation has been driving me crazy for days!
 A: Hint:
If $r$ and $s$ are roots of the quadratic equation $ax^2+bx+c=0$, then $$ax^2+bx+c=a(x-r)(x-s)$$
So
$$3x^2-5x+k=3(x-\tfrac m3)(x-m-3)$$
A: $3x^2 - 5x + k$
Divide equation by 3,
$x^2 - \frac{5x}{3} + \frac k3$
Also given equation have roots $\frac m3$ and $m+3$.
$x^2 - \frac{5x}{3} + \frac k3 = \left(x-\frac m3\right)(x-m-3)$
Now solve right side and compare terms and equate to find m, k.
A: Quadratic equation factorization using values of roots:
$ax^2+bx+c=a(x-x_1)(x-x_2)$
Where $x_1$ and $x_2$ are the roots.
So we get:
$3x^2-5x-k=3(x-\frac{m}{3})(x-m-3)$
$3x^2-5x-k=(3x-m)(x-m-3)$
$3x^2-5x-k=3x^2-3xm-9x-mx+m^2+3m$
$3x^2-5x-k=3x^2-4xm-9x+m^2+3m$
$-5x-k=-4xm-9x+m^2+3m$
$5x+k=4xm+9x-m^2-3m$
$5x+k=(4m+9)x+(-m^2-3m)$
Now by comparing terms, we get:
$\begin{cases}
4m+9=5
\\
-m^2-3m=k
\end{cases}$
$1)\quad 4m+9=5 \\
\hspace{9mm} 4m=-4 \\
\hspace{9mm} m=-1$
$2)\quad -m^2-3m=k \\
\hspace{9mm} -(-1)^2-3\times(-1)=k \\
\hspace{9mm} -1+3=k \\
\hspace{9mm} k=2$
So the answer is:
$\begin{cases}
m=-1 \\
k=2
\end{cases}$
A: Sum of roots of a quadratic equation $= -$( coefficient of $x$ / (coefficient of $x^2$)
So, in your case:
$\frac{5}{3}=m+3+\frac{m}{3}$. This will give you the value of $m$.
Further, 
Product of roots of quadratic equation = constant term / coefficient of $x^2$.
Again, in your case: 
$(m+3)(\frac{m}{3})=k$.
Since you know the value of $m$ from previous equation, you can get $k$ now.
