Question

kaise karu?..........
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if a-b, a+b are the zeros of x^2-4x+3 then (a, b) =

A) (3, 1)
B) (-3,-1)
C) (2, 1)
D) (-2,​

Answer 1

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Answer 2

Explanation:

$f(x) = x {}^{2} - 4x + 3 \\ let \: \: \: one \: \: \: zero \: \: be \: \: \: \ \beta = a - b \\ let \: \: \: other \: \: \: zero \: \: be \: \: \: \gamma = a + b \\ \: \: \: \: where \: \: \: a \: \: \: \: in \: \: \: f(x) = 1 \: \: \: \: \: and \: \: \: b = - 4 \: \: \: and \: \: \: \: c = 3 \\ \beta + \gamma = (a - b) + (a + b) = \frac{ - b}{a} \\ 2a = \frac{ - b}{a} = \frac{4}{1} \\ so \: \: \: \: a = \frac{4}{2} = 2. \: \: \: \: \: \: first \: \: \: \: answer \\ and \: \: \: \: \beta \times \gamma = \frac{c}{a} = \frac{3}{1} \\ (a + b)(a - b) = 3 \\ a {}^{2} - b {}^{2} = 3 \\ 4 - b {}^{2} = 3 \\ - b {}^{2} = - 1 \: \: \: \: \: \: so \: \: \: \: b = 1 \\ so \: \: \: a = 2 \: \: \: \: \: and \: \: \: \: b = 1 \: \: \: \: \: \: \: ans.$

Opt) C is right answer.

Hope it can help u.