d. 3b. 1Maximum nos of zeros in polynomial ax2 bx c.b. 1a. 2c. 4a. 2c.

1.	d. 3b. 1Maximum nos of zeros in polynomial ax2 bx c.b. 1a. 2c. 4a. 2c. 4d. 3
Answer» Multiply both side by a. a^2.x^2+abx+ac=0 (ax)^2+2×ax×b/2+(b/2)^2-b^2/4+ac=0 (ax+b/2)^2=(b^2–4ac)/4 ax+b/2=+/-{(b^2–4ac)^1/2}/2 ax=-b/2+/-{(b^2–4ac)^1/2}/2 x=[-b+/-(b^2–4ac)^1/2]/2a , So the no of zeroes = 2 Like my answer if you find it useful!

d. 3b. 1Maximum nos of zeros in polynomial ax2 bx c.b. 1a. 2c. 4a. 2c. 4d. 3

Answer»

Multiply both side by a.

a^2.x^2+abx+ac=0

(ax)^2+2×ax×b/2+(b/2)^2-b^2/4+ac=0

(ax+b/2)^2=(b^2–4ac)/4

ax+b/2=+/-{(b^2–4ac)^1/2}/2

ax=-b/2+/-{(b^2–4ac)^1/2}/2

x=[-b+/-(b^2–4ac)^1/2]/2a ,

So the no of zeroes = 2

Like my answer if you find it useful!