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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.

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  • $\begingroup$ You have the statement that "Since $\alpha$ is larger than $\beta$, we must have $\alpha= \frac{-b- \sqrt{b^2- 4ac}}{2a}$ and $\beta= \frac{-b+ \sqrt{b^2- 4ac}}{2a}$" but that is backwards. Then from "$\alpha- \beta= 3$, We have $\frac{-b+ \sqrt{b^2- 4ac}}{2a}- \frac{-b- \sqrt{b^2- 4ac}}{2a}= \frac{\sqrt{b^2- 4ac}}{a}= 3$ so that $\sqrt{b^2- 4ac}= 3a$ so that $b^2- 4ac= 9a^2$. But, frankly, I wouldn't do it that way. Instead write $a(x- \alpha)(x- \beta)= ax^2- a(\alpha+ \beta)x+ \alpha\beta= ax^2+ bx+ c$ so we must have $-a(alpha+ \beta)= b$ and $\alpha\beta= c$. $\endgroup$ – user247327 Jul 12 '17 at 18:22
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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

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

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

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  • $\begingroup$ Why do you know to put $(\alpha + \beta)$ as the constant for $bx$ in the quadratic? And why do you know to put $\alpha \beta$ as the constant $c $ in the quadratic? $\endgroup$ – sangstar Jul 12 '17 at 18:17
  • $\begingroup$ @sangstar: because if $\alpha$ and $\beta$ are roots then you can write $(x-\alpha)(x-\beta)=x^2-(\alpha+\beta)+\alpha\beta$ $\endgroup$ – Arnaldo Jul 12 '17 at 18:18
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    $\begingroup$ That's absolutely true... why did I not think of that.. thanks! $\endgroup$ – sangstar Jul 12 '17 at 18:23
  • $\begingroup$ How do we know there's no constant attached to $x^2$? How do we know it's $ax^2$? $\endgroup$ – sangstar Jul 13 '17 at 7:10
  • $\begingroup$ @sangstar: once we have a equation, we can simplify the $a$. $\endgroup$ – Arnaldo Jul 13 '17 at 11:15
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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: $$x^2-9x+18=0 \qquad \text{ and } \qquad x^2+x-2=0.$$

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  • $\begingroup$ Where do you get that $\alpha + \beta = -b$ and $\alpha \beta = c$? $\endgroup$ – sangstar Jul 12 '17 at 18:17
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    $\begingroup$ @sangstar THese are called Viete's relations or a simple way to understand is as follows: Since $x^2+bx+c=(x-\alpha)(x-\beta)=x^2-(\alpha+\beta)x+\alpha\beta$. Now compare the coefficients of both sides. $\endgroup$ – Anurag A Jul 14 '17 at 18:02
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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: $$x^2+x-2=0,\qquad x^2-9x+18.$$

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  • $\begingroup$ What implies the quadratic equation is of the form $x^2-sx+p$ and not $ax^2 \pm sx \pm p$? What implies such specific signage and lack of constant for $x^2$? $\endgroup$ – sangstar Jul 13 '17 at 7:28
  • $\begingroup$ @sangstar: It doesn't imply whatever. This comes from the solution to a standard problem from high school (and actually a theorem): find two numbers, given there sum and their product. If the equation were, say, $ax^2+sx+p=0$, the sum of the roots would be $-s/a$, not $s$, and their product $p/a$. Also, if you factor out $a$ in your equation, you obtain the equation $x^2+\dfrac sa x+\dfrac pa=0$, which is a monic polynomial with the same solutions. $\endgroup$ – Bernard Jul 13 '17 at 9:58
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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,

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

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  • $\begingroup$ @sangstar I have derived $\alpha+\beta = -b/a, \alpha\beta=c/a$ $\endgroup$ – Dhruv Kohli - expiTTp1z0 Jul 12 '17 at 18:24

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