Answer
7 inches
Work Step by Step
The length and the width of the box are $x$
The height is $4$ inches, so it will be $x+4$
Obtain: $f(x)=x(x)(x+4)=x^3+4x^2-63=0$
The leading coefficients: $\pm 1$
The constant terms: $\pm 1, \pm 3,\pm 7, \pm 9,\pm 21,\pm 63$
The possible rational zeros are: $\pm\frac{1}{1},\pm \frac{3}{1},\pm \frac{7}{1},\pm \frac{9}{1},\pm \frac{21}{1},\pm \frac{63}{1}$
Find the solution of the equation: $x=\frac{-b \pm \sqrt b^2-4ac}{2a}=\frac{-7\pm \sqrt (7)^2-4.1.21}{2(1)}=\frac{-7\pm\sqrt -35}{2}=\frac{1\pm i\sqrt 35}{2}$
So, there is no other real zero.
The only real solution of the equation is $x=3$.
The height is $3+4=7$ inches.
