Determining an orthonormal set of basis vectors for the linear space The following is example C.4 from Appendix C (Linear Spaces Review) of Introduction to Laplace Transforms and Fourier Series, Second Edition, by Phil Dyke:

Example C.4 Determine an orthonormal set of vectors for the linear space that consists of all real linear functions: $$\{a+bx:a,b\in\mathbb{R}\ 0\leq x\leq1\}$$ using an inner product $$\langle f,g\rangle=\int_0^1f g\,dx.$$ Solution The set $\{1,x\}$ forms a basis, but is is not orthogonal. Let $a+bx$ and $c+dx$ be two vectors. In order to be orthogonal we must have $$\langle a+bx,c+dx\rangle=\int_0^1(a+bx)(c+dx)\,dx=0.$$ Performing the elementary integration gives the following condition on the constants $a,b,c$ and $d$ $$ac+\frac{1}{2}(bc+ad)+\frac{1}{3}bd=0.$$ In order to be orthonormal too we also need $$\|a+bc\|=1\,\text{ and }\,\|c+dx\|=1$$ and these give, additionally, $$a^2+b^2=1,\ c^2+d^2=1.$$ There are four unknowns and three equations here, so we can make a convenient choice. Let us set $$a=-b=\frac{1}{\sqrt{2}}$$ which gives $$\frac{1}{\sqrt{2}}(1-x)$$
as one vector. The first equation now gives $3c=-d$ from which $$c=\frac{1}{\sqrt{10}},\ \ d=-\frac{3}{\sqrt{10}}.$$ Hence the set $\{(1-x)/\sqrt{10},(1-3x)/\sqrt{10}\}$ is a possible orthonormal one.
  $\ $ $ \ $ Of course there are infinitely many possible orthonormal sets, the above was one simple choice. The next definition follows naturally.

I have the following questions:


*

*How do we determine $a^2 + b^2 = 1$ and $c^2 + d^2 = 1$ from $\|a + bx\| = 1$ and $\|c + dx\| = 1$? This seems similar to the norm of a complex number $a + bi$, but we're not dealing with complex numbers in this case, since we're dealing with the space of all real linear functions, so I'm not sure how these are being derived?

*If we have $a = -b = \dfrac{1}{\sqrt{2}}$ and $3c = -d$, then we have the following: $$\begin{align}
\dfrac{1}{\sqrt{2}}c + \dfrac{1}{2} \left[ \left( \dfrac{-1}{\sqrt{2}} \right)c + \left( \dfrac{1}{\sqrt{2}} \right) (-3c) \right] + \dfrac{1}{3} \left( \dfrac{-1}{\sqrt{2}} \right)(-3c) = 0
\\
\rightarrow \dfrac{c}{\sqrt{2}} - \dfrac{c}{2 \sqrt{2}} - \dfrac{3c}{2\sqrt{2}} + \dfrac{c}{\sqrt{2}} = 0
\\
\rightarrow \dfrac{2c}{\sqrt{2}} - \dfrac{4c}{2\sqrt{2}} = 0
\\
\rightarrow 0 = 0\ ?\end{align}$$
Have I made an error? Where does the $c = \dfrac{1}{\sqrt{10}}$ and $-\dfrac{3}{\sqrt{10}}$ come from?
I would greatly appreciate it if people could please take the time to clarify these.
EDIT: 
The following is proved in the textbook:

Example C.3 Prove that $\|a\|=\sqrt{\langle\mathbf{a}.\mathbf{a}\rangle}\in V$ is indeed a norm for the vector space $V$ with inner product $\langle,\,\rangle$.

Which seems to suggest that $\|x\| = \sqrt{\langle x,x\rangle}$?
 A: By definition of norm induced from inner product, $||x||^2 = \langle x,x\rangle$.
That is, if $||a+bx|| = 1$ then $\langle a+bx,a+bx\rangle = 1$ ,so writing this down:
$$
\int_{0}^1 (a+bx)(a+bx) dx = 1 \implies \int_0^1 (a^2 + 2abx + b^2x^2) dx = 1 \\ \implies a^2 + ab + \frac {b^2}3 = 1
$$ 
therefore, the statement that $a^2 + b^2 = 1$ is FALSE.
This is actually quite clear with an example : $a=0 , b=1$ satisfies $a^2+b^2 = 1$, and gives the polynomial $x$, but $||x||^2 = \int_0^1 x^2 = \frac 13$, so $||x|| \neq 1$. Instead, the other statement given is correct. Replacing $a$ and $b$ by $c$ and $d$ gives you the other analogously correct statement.

Once this happens, you may set $a,b$ to any suitable values, and check what happens to $c$ and $d$. 
For example, set $b = 0$ : from the above equation, this forces $a = \pm 1$, we will take $a =1$.
From the equation that $\langle a+bx,c+dx\rangle = 0$ that the author has derived in your question above, substituting (and cancelling $b$) and rearranging gives $2c+d = 0$, so $d = -2c$.
This must be combined with $c^2 + cd + \frac{d^2}{3} = 1$. Setting it, we get $c^2(1  - 2 + \frac 43) = 1$, so $c^2 = 3$. Just take $c = + \sqrt 3$, so $d = -2\sqrt 3$.
In this manner, we may verify that the polynomial $a+bx = 1$ and $c + dx = \sqrt 3(1 - 2x)$ form an orthonormal basis for the space.
