 minikim38

2022-01-20

A quadratic equation $a{x}^{2}+bx+c=0$ has equal roots at $a=2c$
How could we find the sum of reciprocals of the roots of this equation? Tyrn7i

Expert

Step 1
Since the roots are equal, you must have
${b}^{2}=4ac=8{c}^{2}$
Also, the roots satisfy
$0={x}^{2}+\frac{b}{a}x+\frac{c}{a}$
$={x}^{2}+\frac{b}{a}+\frac{1}{2}$
so the product of the roots is $\frac{1}{2}$; that is, $4{c}^{2}=\frac{1}{2}.$ Therefore, ${b}^{2}=2\left(4{c}^{2}\right)=1$
The sum of the roots is $-\frac{b}{a}$; but it also is equal to 2a. So
$2a=-\frac{b}{a}$
This gives
$2{a}^{2}=-b$,
and we know
$|b|=1$,
so we must have
$b=-1$
So the sum of the reciprocals is
$-\frac{4a}{b}=4a$
Now, ${a}^{2}=\frac{1}{2}$
so either
$a=\frac{\sqrt{2}}{2}$
or
$a=-\frac{\sqrt{2}}{2}$
Thus, the equation is either
$\frac{\sqrt{2}}{2}{x}^{2}-x+\frac{\sqrt{2}}{4}$
or
$-\frac{\sqrt{2}}{2}{x}^{2}-x-\frac{\sqrt{2}}{4}$
Both satisfy the desired conditions: they have a double root at $a=2c$. In one case, the sum of the reciprocals is $2\sqrt{2}$, in the other it is $-2\sqrt{2}$. Jordyn Horne

Expert

Step 1
Also, just for the fun of it, since you have a root with multiplicity two:
$a{x}^{2}+bx+c={\left(\sqrt{ax}+\sqrt{c}\right)}^{2}$
Then the root is
${x}^{\cdot }=±\sqrt{\frac{c}{a}}$
and from $a=2c$, we get
${x}^{\cdot }=±\sqrt{\frac{1}{2}}$
Hence,
$\frac{1}{{x}_{1}}+\frac{1}{{x}_{2}}=\frac{2}{{x}^{\cdot }}=±2\sqrt{2}$ RizerMix

Expert

Step 1This shall answer your question. A polynomial is of the form,$P\left(x\right)=\left(x-{r}_{1}\right)\left(x-{r}_{2}\right)\cdots \left(x-{r}_{n}\right)$where are the roots of the polynomial. Then the first derivative of the polynomial will be,since $\left(x-k{\right)}^{\prime }=1$, we have,$\frac{{P}^{\prime }\left(x\right)}{P\left(x\right)}=\frac{1}{\left(x-{r}_{1}\right)}+\frac{1}{\left(x-{r}_{2}\right)}+\frac{1}{\left(x-{r}_{3}\right)}+\cdots \frac{1}{\left(x-{r}_{n}\right)}$for $x=0$$\frac{{P}^{\prime }\left(0\right)}{P\left(0\right)}=\frac{1}{\left(-{r}_{1}\right)}+\frac{1}{\left(-{r}_{2}\right)}+\frac{1}{\left(-{r}_{3}\right)}+\cdots \frac{1}{\left(-{r}_{n}\right)}$$-\frac{{P}^{\prime }\left(0\right)}{k}=\frac{1}{{r}_{1}}+\frac{1}{{r}_{2}}+\frac{1}{{r}_{3}}+\cdots \frac{1}{{r}_{n}}$where k is the constant term of the polynomial.Therefore the sum of the reciprocals of the roots of any polynomial is equal to minus of the first derivative of the polynomial at $x=0$ divided by the constant term of the polynomial.