Use Mean Value Theorem to establish that $7 \frac{1}{4} < \sqrt{53} < 7 \frac{2}{7}$ I am working on the following question from Lay's Introduction to Analysis with Proof:

Use the Mean Value Theorem to establish that $7 \frac{1}{4} < \sqrt{53} < 7 \frac{2}{7}$

My work so far:
Define $f(t) = \sqrt t$ which we will work with on the interval $[49, 53]$ as $\sqrt{53} \sim \sqrt{49} = 7$.
By MVT,
$$(53-49)\frac{1}{2\sqrt x} = \sqrt{53} - \sqrt{49}$$
Now, $x \in [49, 53]$ so that $49 < x < 53 \Rightarrow 7 < \sqrt x < \sqrt{53} < 8 \Rightarrow \frac{1}{4} < \sqrt{53} - 7 < \frac{2}{7}$
So I have obtained the bound
$$ \frac{1}{4} + 7 < \sqrt{53} < \frac{2}{7} + 7$$
but don't know how to proceed as to  show that $7 \frac{1}{4} < \sqrt{53} < 7 \frac{2}{7}$.

Notice that
$$7\frac{1}{4}=7+\frac{1}{4}$$
and
$$7\frac{2}{7}=7+\frac{2}{7}$$
