Question

Find the equation whose one root is square of the sum and the other root is square of the difference of the roots of the equation 2x² + 2 (m + n) x + m² + n² = 0.

​

Answer 1

Answer:

Question :-

Form the quadratic equation whose roots are the squares of the sum of roots and square

of the difference of roots of the equation

2x² + 2(m+n)x + m² + n² = 0

Answer :-

→ x² - (4mn)x + (m² - n²)² = 0

To Find :-

→ A quadratic equation whose roots are square of the sum of roots and square of the difference of roots of given quadratic.

Solution :-

Let a and b are the roots of required quadratic equation,

→ a = square of the sum of roots of the given polynomial

→ b = square of the difference of roots of the given polynomial.

We have ,

→ 2x² + 2(m+n)x + m² + n² = 0

Hence ,

\begin{gathered} \sf let \: \alpha \: and \: \beta \: are \: the \: roots \: of \: given \: \\ \sf polynomial \\ \to \sf sum \: of \: roots \: = - \frac{coefficient \: of \: x}{ coefficient \: of \: {x}^{2} } \\ \\ \to \sf \alpha + \beta = - \frac{2(m + n)}{2} \\ \\ \to \boxed{ \sf \alpha + \beta \: = - (m + n)} \\ \\ \sf \: product \: of \: roots \: = \frac{constant \: }{ coefficient \: of \: {x}^{2} } \\ \\ \to \boxed{\sf\alpha \beta \: = \frac{ {m}^{2} + {n}^{2} }{2} }\end{gathered}

letαandβaretherootsofgiven

polynomial

→sumofroots=−

coefficientofx

2

coefficientofx

→α+β=−

2

2(m+n)

→

α+β=−(m+n)

productofroots=

coefficientofx

2

constant

→

αβ=

2

m

2

+n

2

Hence ,

\begin{gathered} \to \sf \: a = { (\alpha + \beta)}^{2} \\ \\ \to \sf \: a = { \{ - (m + n) \}}^{2} \\ \\ \to \boxed{ \sf \: a = {(m + n)}^{2} } \\ \\ \bf and \: \\ \to \sf \: b = {( \alpha - \beta)}^{2} \\ \\ \to \sf b = {( \alpha + \beta)}^{2} - 4 \alpha \beta \\ \\ \to \sf \: b = {(m + n)}^{2} - 2( {m}^{2} + {n}^{2} ) \\ \\ \to \sf \: b = {m}^{2} + {n}^{2} + 2mn \: - 2 {m}^{2} - 2 {n}^{2} \\ \\ \to \sf \: \boxed{ \sf \: b = - {(m - n)}^{2} }\end{gathered}

→a=(α+β)

2

→a={−(m+n)}

2

→

a=(m+n)

2

and

→b=(α−β)

2

→b=(α+β)

2

−4αβ

→b=(m+n)

2

−2(m

2

+n

2

)

→b=m

2

+n

2

+2mn−2m

2

−2n

2

→

b=−(m−n)

2

hence , required equation is

→ x² - (a + b)x + ab = 0

\begin{gathered} \sf \: hence \: a + b = {(m + n)}^{2} - {(m - n)}^{2} \\ \\ \to \sf \: a + b = {m}^{2} + {n}^{2} + 2mn - {m}^{2} - {n}^{2} + 2mn \\ \\ \to \boxed{ \sf a + b = 4mn} \\ \\ \sf \: and \: \\ \to \sf ab = - {(m + n)}^{2} {(m - n)}^{2} \\ \\ \to \boxed{\sf \: ab = {( {m}^{2} - {n}^{2} )}^{2} }\end{gathered}

hencea+b=(m+n)

2

−(m−n)

2

→a+b=m

2

+n

2

+2mn−m

2

−n

2

+2mn

→

a+b=4mn

and

→ab=−(m+n)

2

(m−n)

2

→

ab=(m

2

−n

2

)

2

→ x² - (4mn)x + (m² - n²)² = 0

Answer 2

Question :-

Form the quadratic equation whose roots are the squares of the sum of roots and square

of the difference of roots of the equation

2x² + 2(m+n)x + m² + n² = 0

Answer :-

→ x² - (4mn)x + (m² - n²)² = 0

To Find :-

→ A quadratic equation whose roots are square of the sum of roots and square of the difference of roots of given quadratic.

Solution :-

Let a and b are the roots of required quadratic equation,

→ a = square of the sum of roots of the given polynomial

→ b = square of the difference of roots of the given polynomial.

We have ,

→ 2x² + 2(m+n)x + m² + n² = 0

Hence ,

$\sf let \: \alpha \: and \: \beta \: are \: the \: roots \: of \: given \: \\ \sf polynomial \\ \to \sf sum \: of \: roots \: = - \frac{coefficient \: of \: x}{ coefficient \: of \: {x}^{2} } \\ \\ \to \sf \alpha + \beta = - \frac{2(m + n)}{2} \\ \\ \to \boxed{ \sf \alpha + \beta \: = - (m + n)} \\ \\ \sf \: product \: of \: roots \: = \frac{constant \: }{ coefficient \: of \: {x}^{2} } \\ \\ \to \boxed{\sf\alpha \beta \: = \frac{ {m}^{2} + {n}^{2} }{2} }$

Hence ,

$\to \sf \: a = { (\alpha + \beta)}^{2} \\ \\ \to \sf \: a = { \{ - (m + n) \}}^{2} \\ \\ \to \boxed{ \sf \: a = {(m + n)}^{2} } \\ \\ \bf and \: \\ \to \sf \: b = {( \alpha - \beta)}^{2} \\ \\ \to \sf b = {( \alpha + \beta)}^{2} - 4 \alpha \beta \\ \\ \to \sf \: b = {(m + n)}^{2} - 2( {m}^{2} + {n}^{2} ) \\ \\ \to \sf \: b = {m}^{2} + {n}^{2} + 2mn \: - 2 {m}^{2} - 2 {n}^{2} \\ \\ \to \sf \: \boxed{ \sf \: b = - {(m - n)}^{2} }$

hence , required equation is

→ x² - (a + b)x + ab = 0

$\sf \: hence \: a + b = {(m + n)}^{2} - {(m - n)}^{2} \\ \\ \to \sf \: a + b = {m}^{2} + {n}^{2} + 2mn - {m}^{2} - {n}^{2} + 2mn \\ \\ \to \boxed{ \sf a + b = 4mn} \\ \\ \sf \: and \: \\ \to \sf ab = - {(m + n)}^{2} {(m - n)}^{2} \\ \\ \to \boxed{\sf \: ab = {( {m}^{2} - {n}^{2} )}^{2} }$

→ x² - (4mn)x + (m² - n²)² = 0

Hope it helps

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