Question

IF 1 is zero of polynomial p(x)=ax^2 -3(0-1)x-1 then find the value of a
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Answer 1

Heloodude❤️❤️❤️❤️

Explanation

P(x)= ax2 + bx +c = c

ax2+bx = 0

x(ax + b) =0

x=0

or

ax + b = 0

x= -b/a

alpha = 0

beta = -b/a

alpha 2 / beta + beta 2 / alpha== 2 × 0/ -b/a + ( -b/a ÷ 0)

answer is infinity

Hope it's helpful ☺️☺️☺️

Answer 2

Explanation:

$p(x) = a {x}^{2} - 3(0 - 1) x - 1 \: \: \: if \: x = 1 \\ p(1) = a {1}^{2} -3( - 1)x - 1 \\ p(1) = a + 3 \times 1 - 1 \\ p(1) = a + 3 - 1 \\ p(1) = a + 2 \\ now \: put \: eq. \: p(1) = 0 \\ a + 2 = 0 \\ a = - 2$