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प्रश्न
Solve the following equation: `("a+b")^2 "x"^2 - 4 "abx" - ("a - b")^2 = 0`
योग
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उत्तर
`("a+b")^2 "x"^2 - 4 "abx" - ("a - b")^2 = 0`
As, - (a + b)2 + (a - b)2 = - a2 - b2 - 2ab + a2 + b2 - 2ab = - 4ab
(a+b)2x2 -[(a+ b)2-(a-b)2] x - (a - b)2 = 0
(a+ b)2x2 - (a+ b)2x + (a - b)2x - (a - b)2 = 0
{(a+ b)2x} (x - 1) + {(a - b)2} (x - 1) = 0
(x - 1) [(a + b)2x + (a - b)2] = 0
x - 1 = 0 and (a + b)2x + (a - b)2 = 0
x = 1 and x = `-("a" - "b")^2/("a" + "b")^2`
x = 1 and x = `-(("a" - "b")/("a" + "b"))^2`
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