Whakaoti i te left(1+xright)^2*40=135

$${ \left(1+x \right) }^{ 2 } \times 40=135$$

Whakaoti mō x

$$x=\frac{3\sqrt{6}}{4}-1\approx 0.837117307$$

$$x=-\frac{3\sqrt{6}}{4}-1\approx -2.837117307$$

Ngā Upane mā te Kimi Pūtakerua

Whakawehea ngā taha e rua ki te 40.

$$\frac{40\left(x+1\right)^{2}}{40}=\frac{135}{40}$$

Mā te whakawehe ki te 40 ka wetekia te whakareanga ki te 40.

$$\left(x+1\right)^{2}=\frac{135}{40}$$

Whakahekea te hautanga \frac{135}{40} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 5.

$$\left(x+1\right)^{2}=\frac{27}{8}$$

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

$$x+1=\frac{3\sqrt{6}}{4} x+1=-\frac{3\sqrt{6}}{4}$$

Me tango 1 mai i ngā taha e rua o te whārite.

$$x+1-1=\frac{3\sqrt{6}}{4}-1 x+1-1=-\frac{3\sqrt{6}}{4}-1$$

Mā te tango i te 1 i a ia ake anō ka toe ko te 0.

$$x=\frac{3\sqrt{6}}{4}-1 x=-\frac{3\sqrt{6}}{4}-1$$