minikim38

Answered

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?

How could we find the sum of reciprocals of the roots of this equation?

Answer & Explanation

Tyrn7i

Expert

2022-01-21Added 11 answers

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

$\left|b\right|=1$ ,

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 $-2\sqrt{2}$ .

Since the roots are equal, you must have

Also, the roots satisfy

so the product of the roots is

The sum of the roots is

This gives

and we know

so we must have

So the sum of the reciprocals is

Now,

so either

or

Thus, the equation is either

or

Both satisfy the desired conditions: they have a double root at

Jordyn Horne

Expert

2022-01-22Added 16 answers

Step 1

Also, just for the fun of it, since you have a root with multiplicity two:

$a{x}^{2}+bx+c={(\sqrt{ax}+\sqrt{c})}^{2}$

Then the root is

$x}^{\cdot}=\pm \sqrt{\frac{c}{a}$

and from$a=2c$ , we get

$x}^{\cdot}=\pm \sqrt{\frac{1}{2}$

Hence,

$\frac{1}{{x}_{1}}+\frac{1}{{x}_{2}}=\frac{2}{{x}^{\cdot}}=\pm 2\sqrt{2}$

Also, just for the fun of it, since you have a root with multiplicity two:

Then the root is

and from

Hence,

RizerMix

Expert

2022-01-27Added 437 answers

Step 1This shall answer your question. A polynomial is of the form,$P(x)=(x-{r}_{1})(x-{r}_{2})\cdots (x-{r}_{n})$ where ${r}_{1},\text{}{r}_{2},\text{}\cdots \text{}{r}_{n}$ are the roots of the polynomial. Then the first derivative of the polynomial will be,${P}^{\prime}(x)=\frac{(x-{r}_{1}{)}^{\prime}}{(x-{r}_{1})}P(x)+\frac{(x-{r}_{2}{)}^{\prime}}{(x-{r}_{2})}P(x)+\frac{(x-{r}_{1}{)}^{\prime}}{(x-{r}_{1})}P(x)\cdots \text{}\frac{(x-r-n{)}^{\prime}}{(x-{r}_{n})}P(x)$ since $(x-k{)}^{\prime}=1$ , we have,$\frac{{P}^{\prime}(x)}{P(x)}=\frac{1}{(x-{r}_{1})}+\frac{1}{(x-{r}_{2})}+\frac{1}{(x-{r}_{3})}+\cdots \frac{1}{(x-{r}_{n})}$ for $x=0$ $\frac{{P}^{\prime}(0)}{P(0)}=\frac{1}{(-{r}_{1})}+\frac{1}{(-{r}_{2})}+\frac{1}{(-{r}_{3})}+\cdots \frac{1}{(-{r}_{n})}$ $-\frac{{P}^{\prime}(0)}{k}=\frac{1}{{r}_{1}}+\frac{1}{{r}_{2}}+\frac{1}{{r}_{3}}+\cdots \frac{1}{{r}_{n}}$ 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.

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