invertibility of self adjoint operators prove that if $T$ is a self adjoint operator and $a^2$ is less than $4b$ Then $T^2$+$aT$+$bI$ is invertible.
Where $a$ and $b$ are scalars and $I$ is the identity operator
Not: please dont use determinants because the book i am using didn't define determinants yet

This question originally posited the condition $a < 4b$ sufficient for $T^2 + aT + bI$ to be invertible, provided $T$ is self-adjoint; however, I belive that Byron Schmuland was correct to edit this condition to $a^2 < 4b$, and I hope to make my reasons for this clear in what follows, both by proving the corrected proposition and by exhibiting counterexamples to the original one.
Here I shall assume that $T$ is of finite size $n$; that is, that $T$ operates on some vector space $V$ of dimension $n$.  These assumptions are consistent with the approach taken in the second edition of Axler's text, which is the one I have; in a comment, ferrr mentions Axler as his current source. 
These things being said, we observe that the eigenvalues of $T$ are all real, as is well-known:  indeed, if $\lambda$ is an eigenvalue of $T$, then there is a nonzero $v \in V$ with
$Tv = \lambda v ; \tag{1}$
for such $v$ we have
$\lambda \langle v, v \rangle = \langle v, \lambda v \rangle = \langle v, Tv \rangle$ $=\overline{\langle Tv, v \rangle} = \overline{\langle v, T^\dagger v \rangle}$
$=\overline{\langle v, Tv \rangle} = \overline{\langle v, \lambda v \rangle}$
$= \overline {\lambda \langle v, v \rangle} = \bar \lambda \langle v, v \rangle, \tag{2}$
which, since $\langle v, v \rangle \ne 0$, implies that
$\lambda = \bar \lambda; \tag{3}$
note we used $T = T^\dagger$ in establishing (2), (3).  We next invoke the also well-known fact that self-adjoint operators are diagonalizable, or, to put it in more immediate terms, they are possessed of a complete basis of eigenvectors of $V$; that is, there exists a set of $n$ linearly independent vectors $v_i \in V$ such that each satisfies
$Tv_i = \lambda_i v_i \tag{4}$
for some real $\lambda_i$, $1 \le i \le n$; note no $v_i = 0$: we have $V = \text{span}\{v_1, v_2, \ldots, v_n \}$.
From (4) we see that
$T^2 v_i = T(Tv_i) = T(\lambda_i v_i) = \lambda_i T(v_i) = \lambda_i^2 v_i, \tag{5}$
and likewise
$(aT + bI)v_i = (a\lambda_i v_i + b) v_i, \tag{6}$
and hence
$(T^2 + aT + bI) v_i = (\lambda_i^2 + a\lambda_i + b) v_i \tag{7}$
holds for all $v_i$.  We have thus shown that the eigenvalues of $T^2 + aT + bI$ consist of the real numbers $\lambda_i^2 + a\lambda_i + b$, and that the $v_i$ form a complete eigenbasis for the operator $T^2 + aT + bI$.
To proceed further, we examine the function $y: \Bbb R \to \Bbb R$, $y = x^2 + ax + b$.  This is a quadratic polynomial whose first two derivatives are given by
$y'(x) = 2x + a, \tag{8}$
$y''(x) = 2; \tag{9}$
by (8) and (9), $y(x)$ has a global minimum at
$x = -\dfrac{a}{2}, \tag{10}$
and the value of $y$, $y_{\text{min}}$ at this $x$ is
$y_{\text{min}} = (-\dfrac{a}{2})^2 - a(\dfrac{a}{2}) + b = b - \dfrac{a^2}{4}. \tag{11}$
We see that $y_{\text{min}} > 0$ precisely when $a^2 < 4b$ or $(a^2/4) < b$, and since $y(x) \ge y_{\text{min}}$ for all $x \in \Bbb R$, we have shown that every eigenvalue of $T^2 + aT + bI$ is positive.  This in turn implies that $T^2 + aT + bI$ is invertible, since $\ker (T^2 + aT + bI) = \{0\}$ (recall that $\ker (T^2 + aT + bI) = \{0\}$ is spanned by all eigevectors with zero eigenvalue; but are there are none to do the spanning in this case!).  Alternatively, we may explicitly construct $S = (T^2 + aT + bI)^{-1}$ as follows:  since the $v_i$ are linearly independent, we may define $S$ on $\{v_1, v_2, \ldots, v_n\}$ and extend it to $V$ by linearity; we set
$S(v_i) = (\lambda_i^2 + a \lambda_i + b)^{-1}v_i \tag{12}$
for $1 \le i \le n$.  We have
$S(T^2 + aT + bI)(v_i) = (T^2 + aT + bI)S(v_i) = v_i \tag{13}$
for all $v_i$, or
$(T^2 + aT + bI)S = S(T^2 + aT + bI) = I; \tag{14}$
$T^2 + aT + bI$ is thus invertible with inverse $S$.  QED.
Note:  What happens when we hypothesize that $a < 4b$?  Well, if take $1 < a < 4b \le a^2$ these conditions are met, but now $b \le a^2/4$ so $y_{\text{min}} \le 0$, and the parabola $y(x)$ opens upwards; it has real zeroes.  If we define a self-adjoint operator $T_1$ to have eigenvalues at the zeroes of $y(x)$, then $\ker(T_1^2 + aT_1 + bI) \ne 0$, hence there will no inverse for this operator.  So we do need $a^2 < 4b$ to make her fly, Wilbur, for any self-adjoint $T$.  End of Note
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
