Question

If alpha and beta are zeros of x²-k(x+1)-a
And if (alpha+1)(beta+1)=0

Find a...

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Answer 1

Hey friend!!


Here's ur answer...


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$( \alpha + 1)( \beta + 1) = 0 \\ \alpha \beta + \alpha + \beta + 1 = 0 \: \: - - - - -(1)\\ \\ {x}^{2} - k(x + 1) - a = 0 \\ {x }^{2} - kx + (1 - a) = 0\\ \alpha + \beta = \frac{ - b}{a} \\ \\ \alpha + \beta = \frac{k}{1} \\ \alpha + \beta = k. \\ \\ \alpha \beta = \frac{c}{a} \\ \alpha \beta = 1 - a. \\ \\ put \: both \: values \: in \: (1) \\ (1 - a) + k + 1 = 0 \\ 1 - a + k + 1 = 0 \\ 2 - a + k = 0 \\ - a = - 2 - k \\ a = k + 2$

So, value of a = k + 2


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Hope it may help you...


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Answer 2

I think equation is x² - ( k + 1 ) x - a

here ...... ( using coefficient )

a = 1 b = - ( k + 1 ). c = -a


Sum of zeros = - b / a


α + β = ( k + 1 ) / 1

α + β = k + 1 ..................... 1


Product of zeros = c / a

α β = -a / 1

α β = -a .............................. 2





( α + 1 ) ( β + 1 ) = 0

α β + α + β + 1 = 0

-a + k + 1 + 1 = 0

k + 2 - a = 0

k - a = - 2

- a = -2 - k


Sorry but pls verify the question then I will solve and provide u the rest or else I will keep on solving and question will be wrong .