Question

find the quadratic equation whose root are the reciprocals of the root of x^2+4x-10=0

Answer 1

Answer:
$10 {x}^{2} - 4x - 1 = 0$

Solution:

Roots of Quadratic equation

${x}^{2} + 4x - 10 = 0 \\ \\ x_{1,2} = \frac{ - 4 ± \sqrt{16 + 40} }{2} \\ \\ x_{1,2} = \frac{ - 4 ± \sqrt{56} }{2} \\ \\ x_{1,2} = \frac{ - 4 ± 2 \sqrt{14} }{2} \\ \\ x_{1} = - 2 + \sqrt{14} \\ \\ x_{2} = - 2 - \sqrt{14}$
So ,roots of other equation is

$\alpha = \frac{1}{ - 2 + \sqrt{14} } \\ \\ \beta = \frac{1}{ - 2 - \sqrt{14} } \\ \\ \alpha + \beta = \frac{1}{ - 2 + \sqrt{14} } +\frac{1}{ - 2 - \sqrt{14} } \\ \\ = \frac{ - 2 - \sqrt{14} - 2 + \sqrt{14} } {( { - 2)}^{2} - ( { \sqrt{14} )}^{2} } \\ \\ = \frac{ - 4}{4 - 14} \\ \\ \alpha + \beta = \frac{ - b}{a} = \frac{2}{5} \\ \\ \alpha \beta =( \frac{1}{ - 2 + \sqrt{14} }) (\frac{1}{ - 2 - \sqrt{14} } ) \\ \\ \alpha \beta = \frac{c}{a} = \frac{ - 1}{10}$

So if sum and multiplication of zeros are known than polynomial can be find as

${x}^{2} - ( - \frac{b}{a} )x + \frac{c}{a} = 0 \\ \\ {x}^{2} - ( \frac{2}{5} )x + \frac{ - 1}{10} = 0 \\ \\ 10 {x}^{2} - 4x - 1 = 0$

is that polynomial whose zeros are reciprocal of
${x}^{2} + 4x - 10 = 0$
Hope it helps you.

Answer 2

Solution:

Let α and β be the roots of the equation

x²+4x-10=0

sum of the roots:

α+β= -4

product of the roots:

αβ= -10

we have to form the quadratic equation whose roots are 1/α and 1/ β

The required quadratic equation is

$x^2-(\frac{1}{\alpha}+\frac{1}{\beta})x+(\frac{1}{\alpha}.\frac{1}{\beta})=0\\\\x^2-(\frac{\alpha+\beta}{\alpha\beta})x+(\frac{1}{\alpha\beta})=0\\\\x^2-(\frac{-4}{-10})x+(\frac{1}{-10})=0\\\\x^2-(\frac{4}{10})x-(\frac{1}{10})=0\\\\10x^2-4x-1=0$