Question

ax2+bx+c=0 and cx2 +bx+a = 0 have a negative common root then a-b+c=

Answer 1

Lets say that negative root is $- \alpha$

$a \alpha ^{2} - b \alpha + c = 0 \\ c \alpha ^{2} - b \alpha + a = 0 \\ \\ \frac{ \alpha ^{2} }{-ba + bc} = \frac{ \alpha }{ c^{2} - a^{2} } = \frac{1}{-ba + bc} \\ \\ \frac{ \alpha ^{2} }{b(c - a)} = \frac{ \alpha }{(c - a) (c + a) } = \frac{1}{b(c - a)} \\ \\ \frac{ \alpha ^{2} }{b} = \frac{ \alpha }{c + a} = \frac{1}{b} \\ \\ \alpha = \frac{c + a}{b} \\ \alpha ^{2} = 1 \\ \\ \frac{(a + c)}{b} ^{2} = 1 \\ \\ (a + c)^{2} = b^{2} \\ (a + c)^{2} - b^{2} = 0$
(a+c+b) (a+c-b) = 0

Answer
0

Answer 2

a-b+c= 0 ax2+bx+c=0 and cx2 +bx+a = 0 have a negative common root

Step-by-step explanation:

Let say  y is the common root

then ay² + by + c = 0   Eq1

       cy² + by + a = 0   Eq2

Eq1 - Eq 2

=> y²(a - c) + (c - a) = 0

=>  (a - c) (y² - 1) = 0

a ≠ c

because then both equations will be same

& both roots will be same

hence y² - 1 = 0

=> (y + 1)(y - 1) = 0

=> y = -1 or 1

Common roots is -ve

hence y = -1

putting y = -1

a -b + c = 0

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