If alpha, beta are the roots of x² + ax+b=0 then find the equation whose roots are (alpha-beta)² and (alpha + beta)²
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Answers
Answer:
x² - 2(a² - 2b)x + a²(a² - 4b) = 0
Step-by-step explanation:
In an quadratic(ax^2 + bx + c = 0) equation, sum of roots is given by - b/c and product of roots is c/a. Comparing the equations: (let A and B are roots)
Sum of roots = A + B = - a ;
Product of roots = AB = b
Square on both sides of A + B = a:
=> A² + B² + 2AB = a²
=> A² + B² = a² - 2b ...(1)
Square again,
=> A⁴ + B⁴ + 2A²B² = (a² - 2b)²
=> A⁴ + B⁴ + 2A²B² - 4A²B² = (a² - 2b)² - 4A²B²
=> A⁴ + B⁴ - 2A²B² = (a² - 2b)² - 4b²
=> (A² - B²)² = a²(a² - 4b) ...(2)
If there exist an equation with roots (A - B)² and (A + B)²,
Sum of roots = (A - B)² + (A + B)² = 2(A² + B)²
Sum of roots = 2(a² - 2b) [from (1)]
Product of roots = (A - B)²(A + B)² = (A² - B²)²
Product of roots = a²(a² - 4b) [from (2)]
Therefore, required equation is:
=> x² - Sx + P = 0
=> x² - 2(a² - 2b)x + a²(a² - 4b) = 0
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Answer:
x² - 2(a² - 2b)x + a²(a² - 4b) = 0
Step-by-step explanation:
In an quadratic(ax^2 + bx + c = 0) equation, sum of roots is given by - b/c and product of roots is c/a. Comparing the equations: (let A and B are roots)
Sum of roots = A + B = - a ;
Product of roots = AB = b
Square on both sides of A + B = a:
=> A² + B² + 2AB = a²
=> A² + B² = a² - 2b ...(1)
Square again,
=> A⁴ + B⁴ + 2A²B² = (a² - 2b)²
=> A⁴ + B⁴ + 2A²B² - 4A²B² = (a² - 2b)² - 4A²B²
=> A⁴ + B⁴ - 2A²B² = (a² - 2b)² - 4b²
=> (A² - B²)² = a²(a² - 4b) ...(2)
If there exist an equation with roots (A - B)² and (A + B)²,
Sum of roots = (A - B)² + (A + B)² = 2(A² + B)²
Sum of roots = 2(a² - 2b) [from (1)]
Product of roots = (A - B)²(A + B)² = (A² - B²)²
Product of roots = a²(a² - 4b) [from (2)]
Therefore, required equation is:
=> x² - Sx + P = 0
=> x² - 2(a² - 2b)x + a²(a² - 4b) = 0