Question

★ STANDARD QUESTION 5 ★

★ CONTENT QUALITY SUPPORT REQUIRED ★

For what value of " m " will the equation
$\frac{x {}^{2} - bx}{ax - c} = \frac{m - 1}{m + 1}$

have roots equal in magnitude but opposite in sign

★ LEVEL - 100 ★

Answer 1

$things \: to \: know \: \\ 1) \: if \: the \: roots \: of \: the \: quadratic \\ equation \: a {x}^{2} + bx \: + c \: = 0 \\ are \: equal \: in \: magnitude \: and \: \\ opposite \: in \: sign \: then \\ sum \: f \: roots \: is \: 0$
$2)final \: result \: m \: = \frac{a - b}{a + b} and \\ m \: is \: not \: eqal \: to \: - 1$

Answer 2

Solution:

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Given :

$\frac{x^2 -bx}{ax - c} = \frac{m-1}{m+1}$

Roots are equal and opposite

in sign,.

α = -β
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To Find :Value of m,.

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⇒ $\frac{x^2 -bx}{ax - c} = \frac{m-1}{m+1}$⇒ (x² - bx)(m+1) = (m-1)(ax-c)

⇒ (x² - bx)(m + 1) = ( m - 1)(ax - c)

⇒ (m + 1)x² - bmx - bx = amx - mc - ax + c = 0

⇒ (m + 1)x² - bmx - amx + mc + ax - bx - c = 0

⇒ (m+ 1)x² - ((bm + am) - (a - b)) x + mc - c = 0

⇒ (m + 1)x² - (am + bm - a + b ) x + (m - 1)c = 0

⇒ (m + 1)x² - (a(m - 1) + b (m + 1) + (m - 1)c = 0

This equation is in the form,

ax² + bx + c = 0,.

Hence, it is a quadratic equation,.

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Sum of the zeroes = $- \frac{b}{a}$

⇒ α + β = $- \frac{[- (a(m - 1) + b (m + 1)]}{m+1}$

⇒ (-β) + β = $\frac{ (a(m - 1) + b (m + 1)}{m+1}$

⇒ 0 = $\frac{ (a(m - 1) + b (m + 1)}{m+1}$

⇒ 0 = $(a(m - 1) + b (m + 1)$

⇒ 0 = am - a + bm + b

⇒ 0 = am + bm - a + b

⇒ -(-a + b) = am + bm

⇒ a - b = m (a + b)

⇒ $m = \frac{a-b}{a+b}$

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