Whakaoti i te (3x-7)^2-5(2x+1)(x-2)=-x^2-[-(3x+1)]

$$( 3 x - 7 ) ^ { 2 } - 5 ( 2 x + 1 ) ( x - 2 ) = - x ^ { 2 } - [ - ( 3 x + 1 ) ]$$

Whakaoti mō x

$$x = \frac{29}{15} = 1\frac{14}{15} \approx 1.933333333$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(3x-7\right)^{2}.

$$9x^{2}-42x+49-5\left(2x+1\right)\left(x-2\right)=-x^{2}-\left(-\left(3x+1\right)\right)$$

Hei kimi i te tauaro o 3x+1, kimihia te tauaro o ia taurangi.

$$9x^{2}-42x+49-5\left(2x+1\right)\left(x-2\right)=-x^{2}-\left(-3x-1\right)$$

Hei kimi i te tauaro o -3x-1, kimihia te tauaro o ia taurangi.

$$9x^{2}-42x+49-5\left(2x+1\right)\left(x-2\right)=-x^{2}+3x+1$$

Me tāpiri te x^{2} ki ngā taha e rua.

$$9x^{2}-42x+49-5\left(2x+1\right)\left(x-2\right)+x^{2}=3x+1$$

Tangohia te 3x mai i ngā taha e rua.

$$9x^{2}-42x+49-5\left(2x+1\right)\left(x-2\right)+x^{2}-3x=1$$

Whakamahia te āhuatanga tohatoha hei whakarea te -5 ki te 2x+1.

$$9x^{2}-42x+49+\left(-10x-5\right)\left(x-2\right)+x^{2}-3x=1$$

Whakamahia te āhuatanga tuaritanga hei whakarea te -10x-5 ki te x-2 ka whakakotahi i ngā kupu rite.

$$9x^{2}-42x+49-10x^{2}+15x+10+x^{2}-3x=1$$

Pahekotia te 9x^{2} me -10x^{2}, ka -x^{2}.

$$-x^{2}-42x+49+15x+10+x^{2}-3x=1$$

Pahekotia te -42x me 15x, ka -27x.

$$-x^{2}-27x+49+10+x^{2}-3x=1$$

Tāpirihia te 49 ki te 10, ka 59.

$$-x^{2}-27x+59+x^{2}-3x=1$$

Pahekotia te -x^{2} me x^{2}, ka 0.

$$-27x+59-3x=1$$

Pahekotia te -27x me -3x, ka -30x.

$$-30x+59=1$$

Tangohia te 59 mai i ngā taha e rua.

$$-30x=1-59$$

Tangohia te 59 i te 1, ka -58.

$$-30x=-58$$

Whakawehea ngā taha e rua ki te -30.

$$x=\frac{-58}{-30}$$

Whakahekea te hautanga \frac{-58}{-30} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te -2.

$$x=\frac{29}{15}$$