Resolver n+2=sqrt{a-n}

$$n + 2 = \sqrt { a - n }$$

Ebatzi: a

$$a=\left(n+1\right)\left(n+4\right)$$

$$n+2\geq 0$$

Ekuazio irrazionalak ebazteko urratsak

Trukatu aldeak, aldagaiak ezkerraldean egon daitezen.

$$\sqrt{a-n}=n+2$$

Egin ekuazioaren bi aldeak ber bi.

$$a-n=\left(n+2\right)^{2}$$

Egin ken -n ekuazioaren bi aldeetan.

$$a-n-\left(-n\right)=\left(n+2\right)^{2}-\left(-n\right)$$

-n balioari bere burua kenduz gero, 0 da emaitza.

$$a=\left(n+2\right)^{2}-\left(-n\right)$$

Egin -n ken \left(n+2\right)^{2}.

$$a=\left(n+2\right)^{2}+n$$

Ebatzi: a (complex solution)

$$a=\left(n+1\right)\left(n+4\right)$$

$$n=-2\text{ or }arg(n+2)<\pi$$

Ebatzi: n (complex solution)

$$\left\{\begin{matrix}n=\frac{\sqrt{4a+9}-5}{2}\text{, }&a=-2\text{ or }arg(\frac{\sqrt{4a+9}-1}{2})<\pi \\n=\frac{-\sqrt{4a+9}-5}{2}\text{, }&arg(\frac{-\sqrt{4a+9}-1}{2})<\pi \end{matrix}\right.$$

Ebatzi: n

$$n=\frac{\sqrt{4a+9}-5}{2}$$

$$a\geq -2$$