Whakaoti i te sqrt{m+1}+(n-3)^2=0

$$\sqrt { m + 1 } + ( n - 3 ) ^ { 2 } = 0$$

Whakaoti mō m

$$m=-1$$

$$n=3$$

Ngā Upane mō te Whakaoti Whārite Pūtake

Me tango \left(n-3\right)^{2} mai i ngā taha e rua o te whārite.

$$\sqrt{m+1}+\left(n-3\right)^{2}-\left(n-3\right)^{2}=-\left(n-3\right)^{2}$$

Mā te tango i te \left(n-3\right)^{2} i a ia ake anō ka toe ko te 0.

$$\sqrt{m+1}=-\left(n-3\right)^{2}$$

Pūruatia ngā taha e rua o te whārite.

$$m+1=\left(n-3\right)^{4}$$

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

$$m+1-1=\left(n-3\right)^{4}-1$$

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

$$m=\left(n-3\right)^{4}-1$$

Tango 1 mai i \left(n-3\right)^{4}.

$$m=\left(n-4\right)\left(n-2\right)\left(\left(n-3\right)^{2}+1\right)$$

Whakaoti mō n

$$n=3$$

$$m=-1$$