Whakaoti i te y:3(2y+4)=4(2y-1/2)

$$y : 3 ( 2 y + 4 ) = 4 ( 2 y - \frac { 1 } { 2 } )$$

Whakaoti mō y

$$y=\sqrt{22}+5\approx 9.69041576$$

$$y=5-\sqrt{22}\approx 0.30958424$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Me whakarea ngā taha e rua o te whārite ki te 6, arā, te tauraro pātahi he tino iti rawa te kitea o 3,2.

$$2y\left(2y+4\right)=24\left(2y-\frac{1}{2}\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 2y ki te 2y+4.

$$4y^{2}+8y=24\left(2y-\frac{1}{2}\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 24 ki te 2y-\frac{1}{2}.

$$4y^{2}+8y=48y+24\left(-\frac{1}{2}\right)$$

Tuhia te 24\left(-\frac{1}{2}\right) hei hautanga kotahi.

$$4y^{2}+8y=48y+\frac{24\left(-1\right)}{2}$$

Whakareatia te 24 ki te -1, ka -24.

$$4y^{2}+8y=48y+\frac{-24}{2}$$

Whakawehea te -24 ki te 2, kia riro ko -12.

$$4y^{2}+8y=48y-12$$

Tangohia te 48y mai i ngā taha e rua.

$$4y^{2}+8y-48y=-12$$

Pahekotia te 8y me -48y, ka -40y.

$$4y^{2}-40y=-12$$

Me tāpiri te 12 ki ngā taha e rua.

$$4y^{2}-40y+12=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 4 mō a, -40 mō b, me 12 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 4\times 12}}{2\times 4}$$

Pūrua -40.

$$y=\frac{-\left(-40\right)±\sqrt{1600-4\times 4\times 12}}{2\times 4}$$

Whakareatia -4 ki te 4.

$$y=\frac{-\left(-40\right)±\sqrt{1600-16\times 12}}{2\times 4}$$

Whakareatia -16 ki te 12.

$$y=\frac{-\left(-40\right)±\sqrt{1600-192}}{2\times 4}$$

Tāpiri 1600 ki te -192.

$$y=\frac{-\left(-40\right)±\sqrt{1408}}{2\times 4}$$

Tuhia te pūtakerua o te 1408.

$$y=\frac{-\left(-40\right)±8\sqrt{22}}{2\times 4}$$

Ko te tauaro o -40 ko 40.

$$y=\frac{40±8\sqrt{22}}{2\times 4}$$

Whakareatia 2 ki te 4.

$$y=\frac{40±8\sqrt{22}}{8}$$

Nā, me whakaoti te whārite y=\frac{40±8\sqrt{22}}{8} ina he tāpiri te ±. Tāpiri 40 ki te 8\sqrt{22}.

$$y=\frac{8\sqrt{22}+40}{8}$$

Whakawehe 40+8\sqrt{22} ki te 8.

$$y=\sqrt{22}+5$$

Nā, me whakaoti te whārite y=\frac{40±8\sqrt{22}}{8} ina he tango te ±. Tango 8\sqrt{22} mai i 40.

$$y=\frac{40-8\sqrt{22}}{8}$$

Whakawehe 40-8\sqrt{22} ki te 8.

$$y=5-\sqrt{22}$$

Kua oti te whārite te whakatau.

$$y=\sqrt{22}+5 y=5-\sqrt{22}$$

Ngā Upane mō te Whakaoti i te Pūrua

Me whakarea ngā taha e rua o te whārite ki te 6, arā, te tauraro pātahi he tino iti rawa te kitea o 3,2.

$$2y\left(2y+4\right)=24\left(2y-\frac{1}{2}\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 2y ki te 2y+4.

$$4y^{2}+8y=24\left(2y-\frac{1}{2}\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 24 ki te 2y-\frac{1}{2}.

$$4y^{2}+8y=48y+24\left(-\frac{1}{2}\right)$$

Tuhia te 24\left(-\frac{1}{2}\right) hei hautanga kotahi.

$$4y^{2}+8y=48y+\frac{24\left(-1\right)}{2}$$

Whakareatia te 24 ki te -1, ka -24.

$$4y^{2}+8y=48y+\frac{-24}{2}$$

Whakawehea te -24 ki te 2, kia riro ko -12.

$$4y^{2}+8y=48y-12$$

Tangohia te 48y mai i ngā taha e rua.

$$4y^{2}+8y-48y=-12$$

Pahekotia te 8y me -48y, ka -40y.

$$4y^{2}-40y=-12$$

Whakawehea ngā taha e rua ki te 4.

$$\frac{4y^{2}-40y}{4}=-\frac{12}{4}$$

Mā te whakawehe ki te 4 ka wetekia te whakareanga ki te 4.

$$y^{2}+\left(-\frac{40}{4}\right)y=-\frac{12}{4}$$

Whakawehe -40 ki te 4.

$$y^{2}-10y=-\frac{12}{4}$$

Whakawehe -12 ki te 4.

$$y^{2}-10y=-3$$

Whakawehea te -10, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -5. Nā, tāpiria te pūrua o te -5 ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$y^{2}-10y+\left(-5\right)^{2}=-3+\left(-5\right)^{2}$$

Pūrua -5.

$$y^{2}-10y+25=-3+25$$

Tāpiri -3 ki te 25.

$$y^{2}-10y+25=22$$

Tauwehea y^{2}-10y+25. Ko te tikanga pūnoa, ina ko x^{2}+bx+c he pūrua tika pūrua tika pū, ka taea taua mea te tauwehea i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-5\right)^{2}=22$$

Tuhia te pūtakerua o ngā taha e rua o te whārite.

$$\sqrt{\left(y-5\right)^{2}}=\sqrt{22}$$

Whakarūnātia.

$$y-5=\sqrt{22} y-5=-\sqrt{22}$$

Me tāpiri 5 ki ngā taha e rua o te whārite.

$$y=\sqrt{22}+5 y=5-\sqrt{22}$$