The roots of the equation x^2- 3ax + b = 0 differ by 4, then show that 9a^2 = 4b + 16.
Answers
Answer:
Let the roots be = x and y
x - y = 4
Squaring on both sides:
=> (x - y)^2 = 16
=> x^2 + y^2 - 2xy = 16
=> x^2 + y^2 = 16 + 2xy ...(1)
We know that:
x + y = -b/a
[Sum of roots is equal to the ratio of negative coefficient of x is to coefficient of x^2]
Squaring on both sides:
=> (x + y)^2 = (3a)^2
=> x^2 + y^2 + 2xy = 9a^2
Put (1):
=> 16 + 2xy + 2xy = 9a^2
We know that:
xy = c/a
[Product of roots is equal to the ratio of constant is to coefficient of x^2}[/tex]
=> xy = b ....(2)
Put (2) in previous equation:
=> 16 + 4xy = 9a^2
=> 9a^2 = 16 + 4b
proved.
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Answer:
Step by step explanation:
To prove: 9a² = 4b + 16
Given equation:
x² - 3ax + b = 0
Comparing the equation with ax² + bx + c = 0, we get:
a = 1, b = -3a and c = b
Assume the roots as α and β.
Given, Roots of equation differ by 4.
=> α - β = 4
Finding Sum of roots:
α + β = -b/a
= -(-3a) / 1
α + β = 3a
Finding Product of roots:
α β = c/a
= b/1
α β = b
We know that,
(α + β)² = ( α - β )² + 4 ( α β )
=> ( 3a )² = ( 4 )² + 4 ( b )
=> 9a² = 16 + 4b
Hence, Proved.