Math, asked by dipika261019, 10 months ago

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​

Answers

Answered by Sharad001
263

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

Answered by TanmayNema
0

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!

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