Whakaoti i te (b-4)x^2+(2b-8)x+15=0?

$$( b - 4 ) x ^ { 2 } + ( 2 b - 8 ) x + 15 = 0 ?$$

Whakaoti mō b

$$b=-\frac{15-8x-4x^{2}}{x\left(x+2\right)}$$

$$x\neq -2\text{ and }x\neq 0$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te āhuatanga tohatoha hei whakarea te b-4 ki te x^{2}.

$$bx^{2}-4x^{2}+\left(2b-8\right)x+15=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te 2b-8 ki te x.

$$bx^{2}-4x^{2}+2bx-8x+15=0$$

Me tāpiri te 4x^{2} ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

$$bx^{2}+2bx-8x+15=4x^{2}$$

Me tāpiri te 8x ki ngā taha e rua.

$$bx^{2}+2bx+15=4x^{2}+8x$$

Tangohia te 15 mai i ngā taha e rua.

$$bx^{2}+2bx=4x^{2}+8x-15$$

Pahekotia ngā kīanga tau katoa e whai ana i te b.

$$\left(x^{2}+2x\right)b=4x^{2}+8x-15$$

Whakawehea ngā taha e rua ki te x^{2}+2x.

$$\frac{\left(x^{2}+2x\right)b}{x^{2}+2x}=\frac{4x^{2}+8x-15}{x^{2}+2x}$$

Mā te whakawehe ki te x^{2}+2x ka wetekia te whakareanga ki te x^{2}+2x.

$$b=\frac{4x^{2}+8x-15}{x^{2}+2x}$$

Whakawehe 4x^{2}+8x-15 ki te x^{2}+2x.

$$b=\frac{4x^{2}+8x-15}{x\left(x+2\right)}$$

Whakaoti mō x (complex solution)

$$x=\frac{\sqrt{\left(b-19\right)\left(b-4\right)}-b+4}{b-4}$$

$$x=\frac{-\sqrt{\left(b-19\right)\left(b-4\right)}-b+4}{b-4}\text{, }b\neq 4$$

Whakaoti mō x

$$x=\frac{\sqrt{\left(b-19\right)\left(b-4\right)}-b+4}{b-4}$$

$$x=\frac{-\sqrt{\left(b-19\right)\left(b-4\right)}-b+4}{b-4}\text{, }b\geq 19\text{ or }b<4$$