I think I found, with a heavy calculation, a way to obtain such $a,b$.
This answer proves the following claim.
Claim :
For real numbers $s,t,x,y$ such that $ty(x-s)\not=0$, $\color{red}{a=x-s}$ and
$$\color{red}{b^3-( s + 2 x)b^2+s(4 x - s)b-4 s^3 + 13 s^2 x - 15 s x^2 + 5 x^3=0}$$ work, and there is a pair of $(i,j)$ such that
$$a\alpha+b=\frac{s+x}{2}$$
$$a\beta_i+b=s+t\bigg|\frac{x-s}{t}\bigg|\sqrt{10}$$
$$a\gamma_j+b=x+y\bigg|\frac{x-s}{y}\bigg|\sqrt{7}$$
where
$\alpha\approx 1.7419$ is the only one real root of $8x^3+4x^2-10x-37=0$
$\beta$ is a real root of $x^6+4x^5-26x^4-90x^3+200x^2+100x-775=0$ with $\beta_1\approx -1.92038$ and $\beta_2\approx 4.40417$
$\gamma$ is a real root of $x^6-2x^5-22x^4+22x^3+142x^2-272x-131=0$ with $\gamma_1\approx -0.403855$ and $\gamma_2\approx 4.88765$
Examples :
For $(s,t,x,y)=(0,1,1,1)$, we get your example
$$a=1, b^3-2b^2 + 5=0,b\approx -1.2419$$
$$a\alpha+b=\frac{1}{2}$$
$$a\beta_2+b=\sqrt{10}$$
$$a\gamma_2+b=1+\sqrt{7}$$
For $(s,t,x,y)=(0,\color{red}-1,1,\color{red}-1)$, we get a similar example
$$a=1, b^3-2b^2 + 5=0,b\approx -1.2419$$
$$a\alpha+b=\frac{1}{2}$$
$$a\beta_{\color{red}1}+b=\color{red}-\sqrt{10}$$
$$a\gamma_{\color{red}1}+b=1\color{red}-\sqrt{7}$$
For $(s,t,x,y)=(3,1,1,-1)$, we get
$$a=-2, b^3-5b^2+3b-31=0, b\approx 5.4838$$
$$a\alpha+b=2$$
$$a\beta_{\color{red}1}+b=3+2\sqrt{10}$$
$$a\gamma_{\color{red}2}+b=1-2\sqrt{7}$$
In the following, I'm going to prove the claim.
Claim :
For real numbers $s,t,x,y$ such that $ty(x-s)\not=0$, $a=x-s$ and
$$b^3-( s + 2 x)b^2+s(4 x - s)b-4 s^3 + 13 s^2 x - 15 s x^2 + 5 x^3=0$$ work, and there is a pair of $(i,j)$ such that
$$a\alpha+b=\frac{s+x}{2}$$
$$a\beta_i+b=s+t\bigg|\frac{x-s}{t}\bigg|\sqrt{10}$$
$$a\gamma_j+b=x+y\bigg|\frac{x-s}{y}\bigg|\sqrt{7}$$
Proof of Claim :
We have
$$8\alpha^3+4\alpha^2-10\alpha-37=0\tag1$$
$$\beta^6+4\beta^5-26\beta^4-90\beta^3+200\beta^2+100\beta-775=0\tag2$$
$$\gamma^6-2\gamma^5-22\gamma^4+22\gamma^3+142\gamma^2-272\gamma-131=0\tag3$$
Now, let us consider real numbers $a,b$ such that there exist real numbers $p,q,r,s,t,u,x,y,z$ satisfying $a\alpha+b=p+q\sqrt r,a\beta+b=s+t\sqrt u$ and $a\gamma+b=x+y\sqrt z$ where there is no quadratic equation whose root is $b$.
Substituting $\alpha=\frac{p+q\sqrt r-b}{a}$ into $(1)$ gives
$$f_1\sqrt r=f_2$$
where
$$\begin{align}f_1&=A_2b^2+A_1b+A_0
\\f_2&=B_3b^3+B_2b^2+B_1b+B_0
\\A_0&=q(-24p^2-8q^2r-8ap+10a^2)
\\A_1&=q(8a+48p)
\\A_2&=-24q
\\B_0&=37 a^3 + 10 a^2 p - 4 a p^2 - 8 p^3
\\B_1&=-10 a^2 + 8 a p + 24 p^2
\\B_2&=-4a-24p
\\B_3&=8\end{align}$$
Substituting $\beta=\frac{s+t\sqrt u-b}{a}$ into $(2)$ gives
$$f_3\sqrt u=f_4$$
where
$$\begin{align}f_3&=C_5b^5+C_4b^4+C_3b^3+C_2b^2+C_1b+C_0
\\f_4&=D_6b^6+D_5b^5+D_4b^4+D_3b^3+D_2b^2+D_1b+D_0
\\C_0&=100 a^5t + 400 a^4 st - 270a^3 s^2t - 90a^3t^3 u - 104 a^2s^3t - 104 a^2s t^3 u + 20 as^4 t + 40 as^2 t^3 u + 4at^5 u^2 + 6 s^5t + 20 s^3 t^3 u + 6 s t^5 u^2
\\C_1&=-400 a^4 t+ 540 a^3st + 312a^2 s^2 t + 104 a^2 t^3 u - 80 a s^3t - 80 s t^3 u a - 30 s^4t - 60 s^2 t^3 u - 6 t^5 u^2
\\C_2&=-270a^3t - 312 a^2 st + 120 a s^2t + 40 a t^3 u + 60 s^3t + 60 s t^3 u
\\C_3&=104a^2 t - 80 a st - 60 s^2 t - 20 t^3 u
\\C_4&=20 a t + 30 st
\\C_5&=- 6t
\\D_0&=775 a^6 - 100 a^5 s -200 a^4s^2-200a^4 t^2 u + 90a^3s^3 + 270 a^3s t^2 u + 26a^2s^4 + 156a^2 s^2 t^2 u + 26a^2t^4 u^2 - 4a s^5 - 40a s^3 t^2 u - 20 as t^4 u^2 - s^6 - 15 s^4 t^2 u - 15 s^2 t^4 u^2 - t^6 u^3
\\D_1&= 100 a^5 + 400 a^4 s - 270 a^3 s^2 - 270 a^3 t^2 u - 104 a^2 s^3 - 312 a^2 s t^2 u + 20 a s^4 +120 a s^2 t^2 u+ 20 a t^4 u^2 + 6 s^5 + 60 s^3 t^2 u+ 30 s t^4 u^2
\\D_2&=-200 a^4 + 270 a^3 s +156 a^2 s^2 +156 a^2t^2 u - 40 a s^3 - 120 a s t^2 u - 15 s^4 - 90 s^2 t^2 u - 15 t^4 u^2
\\D_3&=- 90 a^3 - 104a^2 s + 40 as^2 + 40a t^2 u + 20 s^3 + 60 s t^2 u
\\D_4&=26 a^2 - 20 a s - 15 s^2 - 15 t^2 u
\\D_5&=4 a + 6 s
\\D_6&=-1\end{align}$$
Substituting $\gamma=\frac{x+y\sqrt z-b}{a}$ into $(3)$ gives
$$f_5\sqrt z=f_6$$
where
$$\begin{align}f_5&=E_5b^5+E_4b^4+E_3b^3+E_2b^2+E_1b+E_0
\\f_6&=F_6b^6+F_5b^5+F_4b^4+F_3b^3+F_2b^2+F_1b+F_0
\\E_5&= - 6y
\\E_4&= - 10 a y + 30 x y
\\E_3&= 88 a^2 y + 40 a x y - 60 x^2 y - 20 y^3 z
\\E_2&= 66 a^3 y - 264 a^2y x - 60 a x^2 y - 20 ay^3 z + 60 x^3 y + 60 x y^3 z
\\E_1&= - 284a^4 y - 132 a^3 x y + 264 a^2y x^2 +88 a^2y^3 z + 40 a x^3 y + 40 a x y^3 z - 30 x^4 y - 60 x^2 y^3 z- 6 y^5 z^2
\\E_0&=-272 a^5 y + 284 a^4x y + 66 a^3x^2 y + 22a^3y^3 z - 88a^2 y x^3 - 88a^2y^3 zx - 10a x^4 y - 20 ax^2 y^3 z - 2ay^5 z^2 + 6 x^5 y + 20 x^3 y^3 z + 6 x y^5 z^2
\\F_6&=-1
\\F_5&= -2 a + 6 x
\\F_4&=22 a^2 + 10 ax - 15 x^2- 15 y^2 z
\\F_3&=22 a^3 - 88 a^2 x -20 a x^2 - 20 a y^2 z + 20 x^3 + 60 x y^2 z
\\F_2&= - 142 a^4 - 66 a^3 x + 132 a^2 x^2 + 132a^2 y^2 z + 20 a x^3 + 60 a x y^2 z - 15 x^4 - 90 x^2 y^2 z - 15 y^4 z^2
\\F_1&=- 272 a^5 + 284a^4 x +66 a^3x^2 + 66 a^3 y^2 z - 88 a^2 x^3 - 264 a^2 x y^2 z - 10a x^4 - 60 a x^2 y^2 z - 10 a y^4 z^2 + 6 x^5 + 60 x^3 y^2 z + 30 x y^4 z^2
\\F_0&=131 a^6 + 272 a^5 x - 142a^4x^2 - 142a^4y^2 z - 22a^3 x^3 - 66 a^3x y^2 z + 22a^2x^4 + 132 a^2x^2 y^2 z + 22a^2y^4 z^2 + 2ax^5 + 20 ax^3 y^2 z + 10 ax y^4 z^2 - x^6 - 15 x^4 y^2 z - 15 x^2 y^4 z^2 - y^6 z^3\end{align}$$
Now, let us consider the case where
$$f_1=f_2=f_3=f_4=f_5=f_6=0$$
Since there is no quadratic equation whose root is $b$, having $f_1=0$, i.e. $A_2b^2+A_1b+A_0=0$ gives $A_2=0$ which implies $q=0$.
Let $K_2b^2+K_1b+K_0$ be the remainder when we divide $f_3$ by $f_2$ where $K$s are represented by $B$s and $C$s.
We see with help of WolframAlpha that $K_0=0$ is equivalent to
$$(a - 2 p + 2 s) (337 a^2 + 60 a p - 60 p^2 + 120 p s - 60 s^2 - 60 t^2 u - 60)=0$$
Considering the case where $a-2p+2s=0$,i.e. $a=2p-2s$, we see that $K_1=0$ is equivalent to
$$(88 p^2 - 176 p s + 88 s^2 - 3 t^2 u) (t^2 u-40 p^2 + 80 p s - 40 s^2 )=0$$
Here, let us consider the case where $t^2 u-40 p^2 + 80 p s - 40 s^2=0$, i.e.
$$u=\frac{40(p-s)^2}{t^2}$$
Similarly, considering the remainder when we divide $f_5$ by $f_2$, we have
$$(2 p - s - x) (184 p^2 - 428 p s + 60 p x + 229 s^2 - 30 s x - 15 x^2 - 15 y^2 z) =0$$
Considering the case where $2p-s-x=0$, i.e. $p=\frac{s + x}{2}$, we also have
$$(13 s^2 - 26 s x + 13 x^2 - 3 y^2 z) (7 s^2 - 14 s x + 7 x^2 - y^2 z)=0$$
Here, let us consider the case where $7 s^2 - 14 s x + 7 x^2 - y^2 z=0$, i.e.
$$z=\frac{7(x-s)^2}{y^2}$$
On the other hand, if
$$\begin{cases}a=2p-2s=s+x-2s=x - s
\\\\u=\dfrac{40(p-s)^2}{t^2}=\dfrac{10 (x - s)^2}{t^2}
\\\\p=\dfrac{s+x}{2}
\\\\z=\dfrac{7(x-s)^2}{y^2}\end{cases}$$
then each of $f_3,f_4,f_5,f_6$ is divisible by $f_2$ as follows :
$$f_3=f_2\bigg(-\frac 34tb^2+\frac{t(s+2x)}{2}b+\frac{t(-39s^2+76sx-40x^2)}{4}\bigg)$$
$$f_4=f_2\bigg(-\frac 18b^3+\frac{s+2x}{8}b^2+\frac{-119s^2+236sx-120x^2}{8}b+\frac{44s^3-13s^2x-105sx^2+75x^3}{8}\bigg)$$
$$f_5=f_2\bigg(-\frac 34yb^2+\frac{y(s+2x)}{2}b+\frac{y(-27s^2+52sx-28x^2)}{4}\bigg)$$
$$f_6=f_2\bigg(-\frac 18b^3+\frac{s+2x}{8}b^2+\frac{-83s^2+164sx-84x^2}{8}b+\frac{32s^3-13s^2x-69sx^2+51x^3}{8}\bigg)$$
Our claim follows from these.$\ \blacksquare$
