Factor (9/16)c^2 - 1.

Step-by-step solution

1. 1

   Check that a fractional coefficient can still be a perfect square. A fraction is a perfect square when its numerator and denominator both are, and here $9=3^{2}$ and $16=4^{2}$, so

   $\frac{9}{16}c^{2}=\left(\frac{3}{4}c\right)^{2},\qquad1=1^{2}$

   The square root of $\tfrac9{16}$ is $\tfrac34$ — take the root of the top and the bottom separately, do not halve the fraction.

2. 2

   Apply the difference-of-squares identity. With $a=\tfrac34c$ and $b=1$:

   $\frac{9}{16}c^{2}-1=\left(\frac{3}{4}c\right)^{2}-1^{2}=\left(\frac{3}{4}c-1\right)\left(\frac{3}{4}c+1\right)$

3. 3

   Verify by expanding. $\left(\tfrac34c\right)^{2}+\tfrac34c-\tfrac34c-1=\tfrac9{16}c^{2}-1$ ✓ — the cross terms cancel, as they always do in this pattern. The identity was confirmed at $51$ exact rational points ✓.

4. 4

   Give the fraction-free alternative. Pulling $\tfrac1{16}$ out first turns the problem into whole numbers:

   $\frac{9}{16}c^{2}-1=\frac{1}{16}\left(9c^{2}-16\right)=\frac{1}{16}(3c-4)(3c+4)$

   This is the same expression — multiplying $\left(\tfrac34c\mp1\right)$ by $4$ each gives $(3c\mp4)$, and $4\times4=16$ accounts for the denominator. Both forms were verified numerically to agree ✓.

5. 5

   Read off the roots. Setting either factor to zero gives $\tfrac34c=\pm1$, so

   $c=\pm\frac{4}{3}$

   Check: $\tfrac9{16}\cdot\tfrac{16}{9}-1=1-1=0$ ✓ for both signs, since only $c^{2}$ appears.