Question

Find the zeroes of the following quadrat
the zeroes and the coefficients.
(1) x² – 2x – 8
(ii) 452
(iv) 4u2 + 8u
Find a quadratic polynomial each with
zeroes respectively.
(v)
-​

Answer 1

Step-by-step explanation:

$(1) \: {x}^{2} - 2x - 8 = 0 \\ {x}^{2} - 4x + 2x - 8 = 0 \\ x(x - 4) + 2(x - 4) = 0 \\ (x - 4)(x + 2) = 0 \\ x - 4 = 0 \: \: x + 2 = 0 \\ x = 4 \: \: \: \: \: x = - 2$

$(4)4 {u}^{2} + 8u = 0 \\ 4( {u}^{2} + 2u ) = 0 \\ {u}^{2} + 2u = 0 \\ u(u + 2) = 0 \\ u = 0 \: \: \: or \: \: \: u + 2 = 0 \\ u = 0 \: \: \: \: or \: \: \: u = - 2$