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 2x^2 + 2(m + n)x + m^2+n^2=0​

Answer 1

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

Answer 2

Answer:

x²-4mnx+(m²-n²)²=0

Formula Used;

(i) α+β=(-b/a)

(ii) α-β=√b²-4ac/|a| ..[From (α-β)²=(α+β)²-4(αβ)]

(iii) QE: x²-(Sum Of Zeroes)x+(Product Of Zeroes)

Here Is Your Solution-

•According To Question;

(i) α and β, Roots Of Given Equation!

(ii) (α+β)² and (α-β)², Roots Of Finding Equation!

(i) We Have; 2x²+2(m+n)x+(m²+n²)=0

• (α+β)=(-b/a)

(α+β)=[-2(m+n)]/2

(α+β)=[-(m+n)]

(α+β)²=(m+n)²

(ii) (α-β)=[√b²-4ac/2a]

(α-β)=[√{2(m+n)}²-4(2)(m²+n²)]/|2|

(α-β)=[√4(m²+n²+2mn)-8(m²+n²)]/2

(α-β)=[√(4m²+4n²+8mn-8m²-8n²)]/2

(α-β)=2[√(-m²-n²+2mn)]/2

(α-β)²= √{-(m-n)²}²

(α-β)²= -(m-n)²

(iii) Now;

Q.E.: x²-[(α+β)²+(α-β)²]x+[(α+β)²(α-β)²]=0

x²-[(m+n)²-(m-n)²]x+[(m+n)(m-n)]²=0

x²-[m²+n²+2mn-m²-n²+2mn]x+[m²-n²]²=0

x²-[4mn]+[m²-n²]²=0

Therefore; x²-4mnx+(m²-n²)²=0

Hope It Helps!