x^2+4kx+(k^2-k+2)=0 шешу

$$x ^ { 2 } + 4 k x + ( k ^ { 2 } - k + 2 ) = 0$$

k мәнін табыңыз (complex solution)

$$k=\frac{\sqrt{12x^{2}-8x-7}}{2}-2x+\frac{1}{2}$$

$$k=-\frac{\sqrt{12x^{2}-8x-7}}{2}-2x+\frac{1}{2}$$

x мәнін табыңыз (complex solution)

$$x=\sqrt{\left(3k-2\right)\left(k+1\right)}-2k$$

$$x=-\sqrt{\left(3k-2\right)\left(k+1\right)}-2k$$

k мәнін табыңыз

$$k=\frac{\sqrt{12x^{2}-8x-7}}{2}-2x+\frac{1}{2}$$

$$k=-\frac{\sqrt{12x^{2}-8x-7}}{2}-2x+\frac{1}{2}\text{, }x\leq -\frac{1}{2}\text{ or }x\geq \frac{7}{6}$$

x мәнін табыңыз

$$x=\sqrt{\left(3k-2\right)\left(k+1\right)}-2k$$

$$x=-\sqrt{\left(3k-2\right)\left(k+1\right)}-2k\text{, }k\leq -1\text{ or }k\geq \frac{2}{3}$$