Question

Find quadratic equation such that its roots are square of sum of the roots and square of difference of the roots of equation: 2x² + 2(p + q)x + p² + q² = 0​

Answer 1

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Answer is in the attachment provided.

Answer 2

Answer:

$\bold\red{ {x}^{2} - (4pq)x - {( {p}^{2} - {q}^{2}) }^{2} = 0}$

Step-by-step explanation:

Given,

a quadratic equation,

$2 {x}^{2} + 2(p + q)x + ( {p}^{2} + {q}^{2} ) = 0$

Now,

To convert it in general form,

divide it with 2,

we get,

$= > {x}^{2} + (p + q)x + \frac{( {p}^{2} + {q}^{2} )}{2} = 0$

Now,

let the roots of the given equation be,$\bold{\alpha \: and \: \beta }$

Therefore,

we have,

Sum of roots,

$\alpha + \beta = - (p + q)$

product of roots,

$\alpha \beta = \frac{( {p}^{2} + {q}^{2} )}{2}$

Therefore,

Square of sum of the roots ,

${( \alpha + \beta )}^{2} = {( - (p + q))}^{2} \\ \\ = {(p + q)}^{2} \\ \\ = {p}^{2} + 2pq + {q}^{2}$

and,

square of Difference of roots,

${( \alpha - \beta )}^{2} = {( \alpha + \beta )}^{2} - 4 \alpha \beta \\ \\ = {(p + q)}^{2} - 4 \times \frac{ {(p}^{2} + {q}^{2} )}{2} \\ \\ = {(p + q)}^{2} - 2( {p}^{2} + {q}^{2} ) \\ \\ = {p}^{2} + {q}^{2} + 2pq - 2 {p}^{2} - 2 {q}^{2} \\ \\ = 2pq - {p}^{2} - {q}^{2}$

Now,

let the ,

square of sum of roots be 'm'

and

square of difference of roots be 'n'

Therefore,

we have,

$m = {p}^{2} + 2pq + {q}^{2} \\ \\ and \\ \\ n = 2pq - {p}^{2} - {q}^{2}$

Therefore,

$m + n = 4pq \\ \\ and \\ \\ mn = - {(p + q)}^{2} {(p - q)}^{2} = - {( {p}^{2} - {q}^{2}) }^{2}$

Now,

the quadratic equation having roots as square of sum of roots i.e, 'm' and square of difference of roots i.e, 'n' is given by,

$\bold{{x}^{2} - (m + n)x + mn = 0}$

Now,

Putting the respective values,

we get,

$= > \bold{ {x}^{2} - (4pq)x - {( {p}^{2} - {q}^{2}) }^{2} = 0}$