You say you need solve
$$0=\alpha+\beta\rho\ (1)$$
$$\alpha^2 + \beta^2=1\ (2)$$
However, I think (2) is improper. You want to get two independent standard normal random variables, which are $X_1$ and $Y=\alpha X_1 + \beta X_2$. $X_1$ is made to follow standard normal. How about $Y$? $Y$ follows normal distribution, the mean of $Y$ is apparently 0, and the variance of $Y$ should be 1, so you need $1=var(Y)=\alpha^2 var(X_1)+\beta^2 var(X_2)+2\alpha\beta cov(X_1,X_2)=\alpha^2 + \beta^2 + 2\alpha\beta\rho$, which means $$1=\alpha^2 + \beta^2 + 2\alpha\beta\rho \ (3)$$ instead of (2).
Therefore, you need solve
$$0=\alpha+\beta\rho\ (4)$$
$$\alpha^2 + \beta^2 + 2\alpha\beta\rho = 1\ (5)$$
From (4) you get $$\alpha=-\beta\rho\ (6)$$
after squaring you have $$\alpha^2=\beta^2\rho^2\ (7)$$
Put (6) and (7) into (5) $$\beta^2\rho^2+\beta^2-2\beta^2\rho^2=1\ (8)$$
$$\beta^2-\beta^2\rho^2=1\ (9)$$
Therefore $$\beta=\pm\sqrt\frac{1}{1-\rho^2}\ (10)$$
Using (6) and (10), you get $$\alpha=\sqrt\frac{\rho^2}{1-\rho^2},\ \beta=-\sqrt\frac{1}{1-\rho^2}$$
or $$\alpha=-\sqrt\frac{\rho^2}{1-\rho^2},\ \beta=\sqrt\frac{1}{1-\rho^2}$$