For a-b-c- R- 0 - let a-b 1-ab -b- b-c 1-bc are in A-P- If - - are the roots of the quadratic equation 2ac x 2 -2abcx-a-c-0- then find the value of 1- 1-

a+b 1 ab ,b, b+c 1 bc are in A.P. 2b=a+b/1 ab+b+c/1 bc ⇒ 2b(1 ab)(1 bc)=(a+b)(1 bc)+(b+c)(1 ab) ⇒ 2b(1 bc ab+ab^2c)=a+b abc b^2x+b+c ab^2 ...

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Standard X

Mathematics

Solving a Quadratic Equation by Factorization Method

For a,b,c∈ ...

Question

For $a, b, c \in R - {0}$ , let $\frac{a + b}{1 - a b}, b, \frac{b + c}{1 - b c}$ are in A.P. If $α, β$ are the roots of the quadratic equation $2 a c x^{2} + 2 a b c x + (a + c) = 0$ , then find the value of $(1 + α) (1 + β)$ .

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Solution

$\frac{a + b}{1 - a b}, b, \frac{b + c}{1 - b c}$ are in A.P.
$⟹ 2 b = \frac{a + b}{1 - a b} + \frac{b + c}{1 - b c}$
$\Rightarrow 2 b (1 - a b) (1 - b c) = (a + b) (1 - b c) + (b + c) (1 - a b)$
$\Rightarrow 2 b (1 - b c - a b + a b^{2} c) = a + b - a b c - b^{2} x + b + c - a b^{2} - a b c$
$\Rightarrow 2 b - 2 b^{2} c - 2 a b^{2} + 2 a b^{3} c = a + 2 b + c - 2 a b c - b^{2} c - a b^{2}$
$\Rightarrow - a - c - b^{2} c - a b^{2} + 2 a b^{3} c + 2 a b c = 0$
$\Rightarrow (a + c) + b^{2} (a + c) - 2 a b c (b^{2} + 1) = 0$
$\Rightarrow (a + c) (1 + b^{2}) - (2 a b c) (1 + b^{2}) = 0$
$\Rightarrow (1 + b^{2}) (a + c - 2 a b c) = 0$
$∵ 1 + b^{2} \neq 0$
$\Rightarrow a + c - 2 a b c = 0$
$\Rightarrow \frac{a + c}{2 a c} = b$
Now, $α$ and $β$ are roots of $(2 a c) x^{2} + (2 a b c) x + (a + c) = 0$
$⟹ α + β = \frac{- 2 a b c}{2 a c} = - b$
$α \times β = \frac{a + c}{2 a c} = b$
$(1 + α) (1 + β) = 1 + (α + β) + α β = 1 - b + b$
$= 1$ .

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For a,b,c∈R−{0}, let a+b1−ab,b,b+c1−bc are in A.P. If α,β are the roots of the quadratic equation 2acx2+2abcx+(a+c)=0, then find the value of (1+α)(1+β).

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