Lutasin ang kz=(2+i)m^2-3(i+1)m-2(1-i)

$$k z = ( 2 + i ) m ^ { 2 } - 3 ( i + 1 ) m - 2 ( 1 - i )$$

I-solve ang k

$$\left\{\begin{matrix}k=-\frac{\left(-2-i\right)m^{2}+\left(3+3i\right)m+\left(2-2i\right)}{z}\text{, }&z\neq 0\\k\in \mathrm{C}\text{, }&\left(m=2\text{ or }m=-\frac{1}{5}+\frac{3}{5}i\right)\text{ and }z=0\end{matrix}\right.$$

Mga Hakbang sa Pagso-solve ng Linear Equation

I-multiply ang 2 at 1-i para makuha ang 2-2i.

$$kz=\left(2+i\right)m^{2}-3\left(i+1\right)m-\left(2-2i\right)$$

Gamitin ang distributive property para i-multiply ang -3 gamit ang i+1.

$$kz=\left(2+i\right)m^{2}+\left(-3i-3\right)m-\left(2-2i\right)$$

Gamitin ang distributive property para i-multiply ang -3i-3 gamit ang m.

$$kz=\left(2+i\right)m^{2}+\left(-3-3i\right)m-\left(2-2i\right)$$

I-multiply ang -1 at 2-2i para makuha ang -2+2i.

$$kz=\left(2+i\right)m^{2}+\left(-3-3i\right)m+\left(-2+2i\right)$$

Ang equation ay nasa standard form.

$$zk=\left(2+i\right)m^{2}+\left(-3-3i\right)m+\left(-2+2i\right)$$

I-divide ang magkabilang dulo ng equation gamit ang z.

$$\frac{zk}{z}=\frac{\left(2+i\right)m^{2}+\left(-3-3i\right)m+\left(-2+2i\right)}{z}$$

Kapag na-divide gamit ang z, ma-a-undo ang multiplication gamit ang z.

$$k=\frac{\left(2+i\right)m^{2}+\left(-3-3i\right)m+\left(-2+2i\right)}{z}$$

I-solve ang m

$$m=\left(\frac{1}{5}-\frac{1}{10}i\right)\sqrt{\left(8+4i\right)kz+\left(24+10i\right)}+\left(\frac{9}{10}+\frac{3}{10}i\right)$$

$$m=\left(-\frac{1}{5}+\frac{1}{10}i\right)\sqrt{\left(8+4i\right)kz+\left(24+10i\right)}+\left(\frac{9}{10}+\frac{3}{10}i\right)$$