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\left(\sqrt{x+1}\right)^{2}=\left(1+\sqrt{4-x}\right)^{2}
Pūruatia ngā taha e rua o te whārite.
x+1=\left(1+\sqrt{4-x}\right)^{2}
Tātaihia te \sqrt{x+1} mā te pū o 2, kia riro ko x+1.
x+1=1+2\sqrt{4-x}+\left(\sqrt{4-x}\right)^{2}
Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(1+\sqrt{4-x}\right)^{2}.
x+1=1+2\sqrt{4-x}+4-x
Tātaihia te \sqrt{4-x} mā te pū o 2, kia riro ko 4-x.
x+1=5+2\sqrt{4-x}-x
Tāpirihia te 1 ki te 4, ka 5.
x+1-\left(5-x\right)=2\sqrt{4-x}
Me tango 5-x mai i ngā taha e rua o te whārite.
x+1-5+x=2\sqrt{4-x}
Hei kimi i te tauaro o 5-x, kimihia te tauaro o ia taurangi.
x-4+x=2\sqrt{4-x}
Tangohia te 5 i te 1, ka -4.
2x-4=2\sqrt{4-x}
Pahekotia te x me x, ka 2x.
\left(2x-4\right)^{2}=\left(2\sqrt{4-x}\right)^{2}
Pūruatia ngā taha e rua o te whārite.
4x^{2}-16x+16=\left(2\sqrt{4-x}\right)^{2}
Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(2x-4\right)^{2}.
4x^{2}-16x+16=2^{2}\left(\sqrt{4-x}\right)^{2}
Whakarohaina te \left(2\sqrt{4-x}\right)^{2}.
4x^{2}-16x+16=4\left(\sqrt{4-x}\right)^{2}
Tātaihia te 2 mā te pū o 2, kia riro ko 4.
4x^{2}-16x+16=4\left(4-x\right)
Tātaihia te \sqrt{4-x} mā te pū o 2, kia riro ko 4-x.
4x^{2}-16x+16=16-4x
Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te 4-x.
4x^{2}-16x+16-16=-4x
Tangohia te 16 mai i ngā taha e rua.
4x^{2}-16x=-4x
Tangohia te 16 i te 16, ka 0.
4x^{2}-16x+4x=0
Me tāpiri te 4x ki ngā taha e rua.
4x^{2}-12x=0
Pahekotia te -16x me 4x, ka -12x.
x\left(4x-12\right)=0
Tauwehea te x.
x=0 x=3
Hei kimi otinga whārite, me whakaoti te x=0 me te 4x-12=0.
\sqrt{0+1}=1+\sqrt{4-0}
Whakakapia te 0 mō te x i te whārite \sqrt{x+1}=1+\sqrt{4-x}.
1=3
Whakarūnātia. Ko te uara x=0 kāore e ngata ana ki te whārite.
\sqrt{3+1}=1+\sqrt{4-3}
Whakakapia te 3 mō te x i te whārite \sqrt{x+1}=1+\sqrt{4-x}.
2=2
Whakarūnātia. Ko te uara x=3 kua ngata te whārite.
x=3
Ko te whārite \sqrt{x+1}=\sqrt{4-x}+1 he rongoā ahurei.