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
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 ,
Hence ,
hence , required equation is
→ x² - (a + b)x + ab = 0
→ x² - (4mn)x + (m² - n²)² = 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!