Question

if one zero of the polynomial p(x)=(k+4)x square + 13x+3k is reciprocal of the other then k is equal to

Answer 1

Hey mate..

Good eve..!!

Here's your solution..

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${(k + 4)x}^{2} + 13x + 3k = 0 \\ \\ let \: zeroes \: are \: \alpha and \: 1 \div \alpha \\ \\ \\ thus \: sum \: of \: zeroes( - b \div a) = \alpha + 1 \div \alpha \\ = > - 13 \div k + 4 = { \alpha }^{2} + 1 \div \alpha \\ \\ product \: of \: zeroes(c \div a) = \alpha \times 1 \div \alpha \\ = > 3 \div k + 4 = 1 \\ 3 = k + 4 \\ - 1 = k$
Thus k will be -1..

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HOPE IT HELPS

@Rêyaañ11