The roots of quadratic equation $3x^2 - 5x - k = 0$ are $m/3$ and $m + 3$. Find the values of $m$ and $k$

Question

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!

Answer 1

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:

$\begin{cases} m=-1 \\ k=2 \end{cases}$

Answer 2

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 $$3x^2-5x+k=3(x-\tfrac m3)(x-m-3)$$

Answer 3

$3x^2 - 5x + k$

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.

Answer 4

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.