the roots of (x-1)(x-3)+k(x-2)(x-4)=0 k>0 are
Answers
The roots of (x - 1)(x - 3) + k(x - 2)(x - 4) = 0 , k > 0 are ...
solution : (x - 1)(x - 3) + k(x - 2)(x - 4) = 0
⇒x² - (1 + 3)x + 3 + k{x² - (2 + 4)x + 2 × 4} = 0
⇒x² - 4x + 3 + kx² - 6kx + 8k = 0
⇒(1 + k)x² - (4 + 6k)x + (3 + 8k) = 0
Discriminant, D = b² - 4ac
= (4 + 6k)² - 4(3 + 8k)(1 + k)
= 16 + 36k² + 48k - 4(8k² + 8k + 3 + 3k)
= 36k² + 48k + 16 - 32k² - 44k - 12
= 4k² + 4k + 4
= 4(k² + k + 1)
but we know, k² + k + 1 > 0 for all real value of k , because D < 0 , a > 0
So, D = 4(k² + k + 1) > 0
Therefore the roots of given equation are real and distinct.
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Answer:-
The roots of (x - 1)(x - 3) + k(x - 2)(x - 4) = 0 , k > 0 are ...
solution :
(x - 1)(x - 3) + k(x - 2)(x - 4) = 0
⇒x² - (1 + 3)x + 3 + k{x² - (2 + 4)x + 2 × 4} = 0
⇒x² - 4x + 3 + kx² - 6kx + 8k = 0
⇒(1 + k)x² - (4 + 6k)x + (3 + 8k) = 0
Discriminant,
D = b² - 4ac
= (4 + 6k)² - 4(3 + 8k)(1 + k)
= 16 + 36k² + 48k - 4(8k² + 8k + 3 + 3k)
= 36k² + 48k + 16 - 32k² - 44k - 12
= 4k² + 4k + 4
= 4(k² + k + 1)
but we know,
k² + k + 1 > 0 for all real value of k , because D < 0 , a > 0
So,
D = 4(k² + k + 1) > 0
Therefore the roots of given equation are real and distinct.
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