Question

Find the zeroes of the quadratic polynomial y^2 - 3y + 2
with the help of the graph.
(a) 1, - 2
(b)-1/4,3/2
(c) 6,- 1
(d) 1,2
I WILL MARK THE CORRECT ANSWER BRIANLY ANS​

Answer 1

Answer:

option d is correct answer

Answer 2

Given  : quadratic polynomial y² - 3y + 2

To Find :  zeroes

(a) 1, - 2

(b)-1/4,3/2

(c) 6,- 1

(d) 1,2

Solution:

polynomial y² - 3y + 2

x = y² - 3y + 2

y = -2

=> x = (-2)² - 3(-2) + 2  = 4 + 6 + 2 = 12

y            x

-2          12

-1           6

0           2

1            0

2           0

3           2

4           6

5          12

Plot the points and draw a  smooth curve passing through the points

Graph Intersect the y axis   at 1 , 2

Hence  Zeroes are 1 , 2

Correct option (d) 1,2

y² - 3y + 2

= y² - y - 2y + 2

= y ( y - 1) - 2( y - 1)

= (y - 1)(y - 2)

Hence zeroes are  1 , 2

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