# Question 445f7

Feb 14, 2018

$5 {x}^{2} - 10 x - 2 = 0$

#### Explanation:

If the equation is $a {x}^{2} + b x + c = 0$ then alpha+beta= b/a and alpha.beta=c/a.
Comparing $a {x}^{2} + b x + c = 0 , {x}^{2} + \left(- 2\right) x + 5 = 0$ we get alpha+beta = -2 and alpha.beta=5.
Now, 1/alpha + 1/beta =(alpha+beta)/alpha.beta= -2/5.
Hence new equation is:
${x}^{2} + \left(- 2\right) x + \left(- \frac{2}{5}\right) = 0$
Or,$5 {x}^{2} - 10 x - 2 = 0$

Feb 14, 2018

see a solution process below;

#### Explanation:

Let $A$ and $B$ be alpha and beta respectively..

${x}^{2} - 2 x + 5 = {x}^{2} - \left(A + B\right) x + A B$

$A + B = 2$

$A B = 5$

When $A + B$ and $\frac{1}{A} + \frac{1}{B}$ are the roots of the equation,

The expression becomes;

x^2-(A+B+(1/A+1/B)x+[(A+B)(1/A+1/B)]

Therefore;

x^2-(A+B+((A+B)/(AB))x+[(A+B)((A+B)/(AB))]#

Substituting the values;

${x}^{2} - \left(2 + \frac{2}{5}\right) x + \left(2 \cdot \frac{2}{5}\right)$

${x}^{2} - \left(\frac{12}{5}\right) x + \frac{4}{5}$

Multiply through by $5$

$5 {x}^{2} - 12 x + 4$

Hope this helps!