Question

find the zeros of the polynomial and verify the relationship between the zeros and coefficients a) 10y^2 3y-1. b) x^2 1/6x-2

Answer 1

the discriminant of the quadratic equations 2x+5x+k=0 then find the Value of k.

Answer 2

Step-by-step explanation:

$(a) :10 {y}^{2} +3y-1\\10 {y}^{2} +5y-2y-1=0\\5y(2y+1)-1(2y+1)=0\\(5y-1)(2y+1)=0\\y= \frac{1}{5} \:,\: \frac{-1}{2} \\ \: \: \: \: \: \: α\: \: ,\: \: β \\ Sum \: of \: zeroes: \\ α+β=- \frac{b}{c} \\ \frac{1}{5} + \frac{(-1)}{2} = - \frac{3}{10} \\ \frac{2 - 5}{10} = - \frac{3}{10} \\ - \frac{3}{10} = - \frac{3}{10} \: \: \: verified \\ product \: of \: zeroes: \\ \alpha \beta = \frac{c}{a} \\ \frac{1}{5} \times \frac{ - 1}{2} = \frac{ - 1}{10} \\ \frac{ - 1}{10} = \frac{ - 1}{10} \: \: verified \\ (b) : {x}^{2} + \frac{1}{6} x-2\\ \frac{6 {x}^{2} + x - 12}{6} =0\\ 6 {x}^{2} + x - 12 =0\\ 6 {x}^{2} + 9x - 8x - 12 = 0 \\ 3x(2x+3)-4(2y+3)=0\\(3x-4)(2x+3)=0\\x= \frac{4}{3} \:,\: \frac{ - 3}{2} \\ \: \: \: \: \: \: α\: \: ,\: \: β \\ Sum \: of \: zeroes: \\ α+β=- \frac{b}{c} \\ \frac{4}{3} + \frac{(-3)}{2} = - \frac{1}{6} \\ \frac{8 - 9}{6} = - \frac{1}{6} \\ - \frac{1}{6} = - \frac{1}{6} \: \: \: verified \\ product \: of \: zeroes: \\ \alpha \beta = \frac{c}{a} \\ \frac{4}{3} \times \frac{ - 3}{2} = \frac{ - 12}{6} \\ - 2 = - 2 \: \: verified$

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