Question

8y^2-3y^2 find the zeroes of th polynomial​

Answer 1

Answer:

x=8y−3 -- (1)

8y=x+3

y=

8

x+3

From equation (1) assume the value of x and y to satisfy the equation to zero.

Plotting (−3,0),(0,0.375) and joining them, we get another straight line.

The graph of x=8y−3 intersect the x-axis at (−3,0). Hence the zero of x=8y−3 is (−3,0) as it intersects at x-axis at (−3,0).

solution