# I have a problem solving a system of equations. The problem is that I don't know how to continue the process.

I need help with a problem based on system of equations:

$$\left\{ \begin{array}{c} -1=(-b-\sqrt{b^2-4ac})/2a \\ 3=(-b+\sqrt{b^2-4ac})/2a \\ \end{array} \right.$$

I tried to solve it and I arrived here: $$\left\{ \begin{array}{c} a=-b/2 \\ \sqrt{b^2+2bc}=-2b \\ \end{array} \right.$$

then, I don't know how to continue to solve it, I tried to square both sides but It seems that it doesn't give the exact result.

## 4 Answers

note that what you have is: $$\alpha=\frac{-b-\sqrt{b^2-4ac}}{2a}=-1$$ $$\beta=\frac{-b+\sqrt{b^2-4ac}}{2a}=3$$ and we know that the quadratic formula, solving the general equation: $$ax^2+bx+c=0$$ is given as: $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ so we can also say that: $$f(x)=(x-\alpha)(x-\beta)=0$$ $$(x+1)(x-3)=0$$ $$x^2-2x-3=0$$ so:

$$a=1$$, $$b=-2$$ and $$c=-3$$

• While I like the elegance of this approach, you found not all solutions. If you scale $a$, $b$ and $c$ with a common non-zero factor, this will give you another solution. – M. Winter Oct 11 '18 at 6:40

Let $$x_1=\frac{-b-\sqrt{b^2-4ac}}{2a}$$ and $$x_2=\frac{-b+\sqrt{b^2-4ac}}{2a}.$$

Thus, $$x_1+x_2=2$$ and $$x_1x_2=-3,$$ which says that $$x_1$$ and $$x_2$$ they are roots of the eqation $$x^2-2x-3=0$$ and also $$ax^2+bx+c=0,$$ where $$a\neq0$$ by the given.

Finally, $$b=-2a$$ and $$c=-3a$$.

Adding the two equations, we get $$\frac{-b}{a}=2.$$ Multiplying the two equations, we get $$\frac{c}{a}=-3.$$ So for any non-zero $$a$$, we have $$b=-2a$$ and $$c=-3a$$.

• Adding the two equations and multiplying them? I don't understand what you mean. – Norman Facco Oct 2 '18 at 14:04
• @NormanFacco it is exactly what everyone else has done. There are two equations, when you add them, the square root terms cancel out and you are left with only $-b/a$ on one side and $3-1=2$ on the other side. Now start again and multiply the two equations. – Anurag A Oct 2 '18 at 20:31

It might be that you were asked to find expressions for $$b,c$$ in $$a$$ (or something like that) under the condition that the quadratic equation $$ax^2+bx+c=0$$ has $$-1$$ and $$3$$ as roots.

If so then realize that: $$ax^2+bx+c=a(x-3)(x+1)=ax^2-2ax-3a$$telling you that $$b=-2a$$ and $$c=-3a$$.

Found is actually that: $$b:c:a=(-2):(-3):1$$

• Thanks for the answer, just two things I want to ask: 1) so a can be also, for example, 2? 2) b:c:a what does it mean? – Norman Facco Oct 2 '18 at 12:14
• If you must solve $(a,b,c)$ then there are the possibilities $(a,-2a,-3a)$ where $a\neq0$. So indeed you can have $a=2$. $b:c:a$ is a notation concerning the ratios. Unessential. The equality just says here: if $a=1$ then $b=-2$ and $c=-3$. – drhab Oct 2 '18 at 12:36
• You helped me a lot! New things learnt today! :) – Norman Facco Oct 2 '18 at 12:59
• You are welcome. – drhab Oct 2 '18 at 13:17