Express $c$ and $d$ in terms of $m$ where $c$ and $d$ are zeroes of $f$ where $m > -2$

Question

Let $$f(x) = x^2 - mx -(6m^2+25m+25)$$ where $m > - 2$

It can be shown that $f(x)$ has two zeroes.

Suppose we have $c,d \in \mathbb R$ s.t. $c < d$ and $f(c) = f(d) = 0$, express $c$ and $d$ in terms of $m$, again > -2.

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What I tried:

$c,d$ are the zeroes of $f$ right?

If so then

$$cd = -(6m^2+25m+25)$$

and

$$c+d = m$$

I solved the following:

$$d = \frac{m + \sqrt{m^2 + 4(1)(6m^2-25m-25)}}{2(1)}$$

$$c = \frac{m - \sqrt{m^2 + 4(1)(6m^2-25m-25)}}{2(1)}$$

Is that right?

Answer 1

Since you have a quadratic formula of the form $$ax^2+bx+c=0$$ The well know quadratic formula $$x=\frac {-b\pm\sqrt {b^2-4ac}}{2a}$$ Gives us our solutions for $x $. Applying this we find $$\begin{align} x &=\frac {m\pm\sqrt{m^2+4 (6m^2+25m+25)}}{2} \\ &= \frac {m\pm\sqrt{25(m+2)^2}}{2} \\ &= \frac {m\pm5(m+2)}{2} \end{align}$$ Because $m>-2$ the discriminant is positive so we have that $d $ is given by choosing the $+$ and $c $ by choosing the $-$.