Question

(z+3/4)(z+4/3) please solve this.

Answer 1

$(z + \frac{3}{4} )(z + \frac{4}{3} ) \\ \\ = z(z + \frac{4}{3} ) + \frac{3}{4} (z + \frac{4}{3} ) \\ \\ = {z}^{2} + \frac{4z}{3} + \frac{3z}{4} + 1 \\ \\ = {z}^{2} + \frac{16z + 9z}{12} + 1 \\ \\ = {z}^{2} + \frac{25z}{12} + 1 \\ \\ = 12( {z}^{2} + \frac{25z}{12} + 1) \\ \\ = 12 {z}^{2} + 25z + 12 \\ \\ = 12 {z}^{2} + 16z + 9z + 12 \\ \\ = 4z(3z + 4) + 3(3z + 4) \\ \\ = (4z + 3)(3z + 4) \\ \\ 4z + 3 = 0 \\ 4z = - 3 \\ z = - \frac{3}{4} \\ \\ 3z + 4 = 0 \\ 3z = - 4 \\ z = - \frac{4}{3} \\ \\ \bf \: So, \: the \: value \: of \: z \: is \: either \: - \frac{3}{4} \\ \bf \: or \: - \frac{4}{3}$

If any doubt, please ask ;)

Answer 2

$\underline \bold {Solution:-}$

$(z + \frac{3}{4} )(z + \frac{4}{3} ) \\ \\ on \: multiplying \: them \\ \\ = z(z + \frac{4}{3} ) + \frac{3}{4} (z + \frac{4}{3} ) \\ \\ on \: multiplying \: inside \: the \: bracket \\ \\ = {z}^{2} + \frac{4z}{3} + \frac{3z}{4} + \frac{3}{4} \times \frac{4}{3} \\ \\ = {z}^{2} + \frac{4z}{3} + \frac{3z}{4} + 1 \\ \\ on \: taking \: the \: lcm \: of \: middle \: term \\ \\ = {z}^{2} + \frac{4 \times 4z + 3 \times 3z}{3 \times 4} + 1 \\ \\ = {z}^{2} + \frac{16z + 9z}{12} + 1 \\ \\ = {z}^{2} + \frac{25z}{12} + 1\\ \\ = \frac{1}{12} (12 {z}^{2} + 25z + 12)$

It can be solved only this much.

It's form can be changed further , but....


The value of z cannot be find out from it , because the information given is not sufficient for that