Question

Find the equation whose roots are √2, -√2 , 3i-3i i
s

Answer 1

Step-by-step explanation:

The equation is

$(x - \sqrt{2} )(x - ( - \sqrt{2} ))(x - 3i)(x - (3i)) = 0$

$( {x}^{2} - {(\sqrt{2} )}^{2} )( {x}^{2} - {(3i)}^{2} ) = 0$

$( {x}^{2} - 2)( {x}^{2} - 9 {i}^{2} ) = 0$

$( {x}^{2} - 2)( {x}^{2} - 9( - 1)) = 0$

$( {x}^{2} - 2)( {x}^{2} + 9 ) = 0$

${x}^{4} + 9 {x}^{2} - 2 {x}^{2} - 18 = 0$

${x}^{4} + 7 {x}^{2} - 18 = 0$

Answer 2

The required equation is x⁴ + 7x² - 18 = 0

Correct question : Find the equation whose roots are √2 , - √2 , 3i , - 3i

Given :

For a equation roots are √2 , - √2 , 3i , - 3i

To find :

The equation

Solution :

Step 1 of 2 :

Write down the given roots

Here it is given that for a equation roots are √2 , - √2 , 3i , - 3i

Step 2 of 2 :

Find the equation

The required equation is given by

$\displaystyle \sf{ (x - \sqrt{2} ) \{x - ( - \sqrt{2}) \}(x - 3i) \{x - ( - 3i) \} = 0 }$

$\displaystyle \sf{ \implies (x - \sqrt{2} ) (x + \sqrt{2})(x - 3i) (x + 3i) = 0 }$

$\displaystyle \sf{ \implies (x + \sqrt{2} ) (x - \sqrt{2})(x + 3i) (x - 3i) = 0 }$

$\displaystyle \sf \implies \{ {(x)}^{2} - {( \sqrt{2} )}^{2} \} \{ {(x)}^{2} - {(3i)}^{2} \} = 0\: \: \: \bigg[ \: \because \: a^2 - b^2 = (a + b) (a-b) \bigg]$

$\displaystyle \sf{ \implies ( {x}^{2} - 2)( {x}^{2} - 9 {i}^{2} ) = 0 }$

$\displaystyle \sf{ \implies ( {x}^{2} - 2)( {x}^{2} + 9 ) = 0 }\: \: \: \bigg[ \: \because \: {i}^{2} = - 1 \bigg]$

$\displaystyle \sf{ \implies {x}^{2}( {x}^{2} + 9 ) - 2( {x}^{2} + 9 ) = 0 }$

$\displaystyle \sf{ \implies {x}^{4} + 9 {x}^{2} - 2 {x}^{2} - 18 = 0 }$

$\displaystyle \sf{ \implies {x}^{4} + 7 {x}^{2} - 18 = 0 }$

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