Finding possible quadratic equations given conditions of their roots

Question

I have an issue with the following problem:

Two quadratic equations have real roots $\alpha$ and $\beta$ such that $$\alpha - \beta = 3$$ and $$\alpha \beta = 2(\alpha + \beta).$$ Find the two possible quadratic equations that satisfy these conditions.

Since $\alpha$ is larger than $\beta$, in the general solution to the quadratic equation, familiarly:

$$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

$\alpha$ needs to be the larger solution so it must be of the form:

$$\alpha = \frac{-b - \sqrt{b^2-4ac}}{2a}$$

since it's the bigger root, and $\beta$ is therefore:

$$\beta = \frac{-b + \sqrt{b^2-4ac}}{2a}$$

However, I need to find the two quadratic equations that have these roots.

If I play around a bit with the properties of the roots, I can work out some things like:

$$\alpha - \beta = 3$$

$$\frac{1}{2a}\ ((-b - \sqrt{b^2-4ac}) - (-b + \sqrt{b^2-4ac}) = 3$$

$$ - 2\sqrt{b^2-4ac} = 6a$$

$$4b^2 - 16ac = 36a^2$$

$$b = \frac{\sqrt{16a(2a - c)}}{2}$$

And the next property:

$$\alpha \beta = 2(\alpha + \beta)$$

$$\frac{-b - \sqrt{b^2-4ac}}{2a} \bullet \frac{-b + \sqrt{b^2-4ac}}{2a} = 2(\frac{1}{2a}(-b - \sqrt{b^2-4ac}+ -b + \sqrt{b^2-4ac}))$$

$$\frac{4c}{4a}=\frac{-2b}{a}$$

$$4c=\sqrt{16a(2a - c)}$$

$$16c^2 = 32a^2 - 16ac$$

$$0= 2a^2-ac -c^2$$

$$(a-c)(2a+c)=0$$

$$a= c$$ $$2a = -c$$

I'm going to assume these will represent both quadratics, so I'll use $a = c$ first.

$$b = \frac{\sqrt{16a(2a - c)}}{2}$$

$$b = \frac{\sqrt{16c(2c - c)}}{2}$$

$$b = \frac{\sqrt{16c^2}}{2}$$ $$b= \pm\ 2c$$

I'll use $b = 2c$ for equation one.

$$\frac{1}{2a}\ ((-b - \sqrt{b^2-4ac}) - (-b + \sqrt{b^2-4ac}) = 3$$

$$\frac{1}{2c}\ ((-2c - \sqrt{{4c}^2-4c^2}) - (-2c + \sqrt{{4c}^2-4c^2}) = 3$$

$$c=0$$

Obviously, I blew it, since that implies $a = b = c$. I'm not sure if what I did wrong was some technical errors or the wrong approach in itself. I'd appreciate any guidance on this.

Answer 1

The quadratic equation has the form

$$x^2-(\alpha+\beta)x+\alpha\beta=0$$

but $\alpha=3+\beta$ and

$$\alpha\beta=2(\alpha+\beta)\to (3+\beta)\beta=2(3+2\beta)$$

$$\beta^2-\beta-6=0\to \beta=-2 \text{ or }\beta=3$$

which give us $\alpha=1$ or $\alpha =6$, so we get the equations

$$x^2-(1-2)x+(1)(-2)=0\to x^2+x-2=0$$

$$x^2-(6+3)x+(6)(3)=0\to x^2-9x+18=0$$

Answer 2

Suppose the quadratic equation whose roots are $\alpha, \beta$ is $x^2+bx+c=0$, then $$\alpha+\beta=-b\qquad \text{and} \qquad \alpha\beta=c.$$

So we need $\alpha+\beta$ and $\alpha\beta$

\begin{align*} (\alpha+\beta)^2 & = (\alpha-\beta)^2+4\alpha\beta\\ & = 9+8(\alpha+\beta)\\ t^2-8t-9&=0 && (\text{assume } t=\alpha+\beta)\\ t&=9,-1 \end{align*} So $$\alpha+\beta=9,-1 \qquad \text{ and } \qquad \alpha\beta=18,-2$$ So the two possible quadratics are: $$x^2-9x+18=0 \qquad \text{ and } \qquad x^2+x-2=0.$$

Answer 3

Let $\alpha$ and $\beta$ be the roots of the quadratic equation,

$$ax^2+bx+c=0$$

Then,

$$\alpha+\beta = -b/a \ \text{ and } \ \alpha\beta = c/a$$

Note: You can get these by,

$$\alpha+\beta = \frac{-b+\sqrt{b^2-4ac}}{2a} + \frac{-b-\sqrt{b^2-4ac}}{2a} = \frac{-b}{a}$$

$$\alpha\beta = \frac{(-b)^2 - (b^2-4ac)}{4a^2} = \frac{c}{a}$$

Now, using given conditions,

$$(\alpha - \beta)^2 = \alpha^2 + \beta^2 - 2\alpha\beta = 9 \\ \implies (\alpha+\beta)^2-4\alpha\beta = 9 \implies \alpha\beta^2 - 16\alpha\beta = 36 \implies \alpha\beta = \frac{16 \pm \sqrt{16^2+4\cdot 36}}{2}$$

$$\implies \alpha\beta = 8\pm10 = 18 \text{ or } -2$$

Now, use $\alpha+\beta = \alpha\beta/2 = 9 \text{ or } -1$

So, two quadratic equations are,

$$x^2 -9x+18= 0 \ \ \text{ and } \ \ x^2 +x -2 = 0$$

Answer 4

It's much simpler to solve using Vieta's relations. Denote $s=\alpha+\beta$, $p=\alpha\beta$. Remember from high school that, if we know the values of $s$ and $p$, $\alpha$ and $\beta$ are the roots of the quadratic equation: $$x^2-sx+p=0.$$

The second condition becomes $p=2s$. As to the first, it is equivalent to $$(\alpha-\beta)^2=9\iff (\alpha+\beta)^2-4\alpha\beta=s^2-4p=9.$$ Taking into account the second relation, we obtain the quadratic equation $$s^2-8s-9=(s-4)^2-25=0,\quad\text{whence}\quad s=4\pm5=-1,9,\enspace p=-2, 18,$$ and the two quadratic equations: $$x^2+x-2=0,\qquad x^2-9x+18.$$