Question

if a & ß are the zeroes of the polynomial x^2-3x+2 then find 1/a+1/ß​

Answer 1

Answer:

${x}^{2} - 3x + 2 = 0 \\ {x }^{2} - 2x - x + 2 = 0 \\ x(x - 2) - 1(x - 2) = 0 \\ (x - 2)(x - 1) = 0 \\ x -2 = 0 \: \: \: \: \: \: \: \: x - 1 = 0 \\ x = 2 = a \: \: \: \: \: \: \: \: \: \: \: \: \: \: x = 1 = b \\ now.. \\ \frac{1}{a} + \frac{1}{b} = \frac{1}{2} + \frac{1}{1} = \frac{1 + 2}{2} = \frac{3}{2} ans.$

Answer 2

Answer:

hey mate here us yur ans

alpha + beta = -b/a = -(-3)/1 = 3

alpha.beta = c/a = 2/1 = 2

alpha.(3-alpha)= 2

3alpha-(alpha)² = 2

(alpha)²-3alpha+2 = 0

(alpha)²-2alpha-alpha+2 = 03-1 = 2

alpha(alpha-2)-1(alpha-2)=0

(alpha-1).(alpha-2)

if alpha -1 = 0 =>alpha= 1 so, beta = 2

if alpha-2 = 0 => alpha = 2 so, beta = 1

then

case 1: alpha = 1 beta = 2

1/1+1/2 = 3/2

case 2: alpha = 2 beta = 1

1/2+1/1 = 3/2

hope it help u