1.13 Rearranging formulas and radical equations

An established formula may need to be rewritten so that the quantity of interest can be calculated directly.

Equivalent formula. A rearranged equation that expresses the same relationship while isolating a different variable.

To solve a formula for one variable, treat the remaining symbols as constants. From

\[S=2\pi rh+2\pi r^2,\]

subtract \(2\pi r^2\) and divide by \(2\pi r\):

\[h=\frac{S-2\pi r^2}{2\pi r},\qquad r\ne0.\]

Worked Example: Isolating height in a cylinder formula

Starting with \(S=2\pi rh+2\pi r^2\), subtract \(2\pi r^2\) and divide by \(2\pi r\) to obtain

\[h=\frac{S-2\pi r^2}{2\pi r},\qquad r\ne0.\]

Try It. Solve \(Q=Av\) for \(v\).
Check Your Work \(v=Q/A\), with \(A\ne0\).

Worked Example: Rearranging the hydraulic residence-time formula

Hydraulic residence time is \(t=V/Q\). Solving for flow rate gives

\[tQ=V\qquad\Rightarrow\qquad Q=\frac Vt,\qquad t\ne0.\]

Try It. Rearrange \(L=Qt\) to solve for \(t\).
Check Your Work \(t=L/Q\), with \(Q\ne0\).

Rearranging formulas uses the same equivalent operations as solving numerical equations. Treat all other symbols as constants. If the desired variable occurs in several terms, collect those terms and factor the variable before dividing. Record restrictions created by division, and check the new formula by substituting it into the original relationship.

For a radical equation, determine the domain, isolate the radical, raise both sides to the required power, solve, and check every candidate in the original equation. If another radical remains, isolate it and repeat. The check is essential because squaring can introduce extraneous solutions.

1.13.1 Formula and radical checks

If the desired variable appears in several terms, collect those terms and factor the variable before dividing. For example, \(F=GmM/r^2\) becomes \(Fr^2=GmM\), then \(m=Fr^2/(GM)\). The final formula should have the units expected for \(m\).

For a radical equation, the domain can eliminate candidates before substitution. In \(\sqrt{x+2}=x-4\), the nonnegative left side requires \(x\ge4\). Any solution produced after squaring that is below 4 is extraneous.

Worked Example: Rejecting an extraneous solution

For \(\sqrt{x+5}=x-1\), the right side requires \(x\ge1\). Squaring gives \(x+5=x^2-2x+1\), so \((x-4)(x+1)=0\). Only \(x=4\) satisfies the original equation.

Try It. Solve \(\sqrt{x+6}=x\).
Check Your Work Squaring gives \(x=3\) or \(-2\); only \(x=3\) checks.

Worked Example: Solving a radical depth equation

Solve \(\sqrt{2d+3}=d\). The right side requires \(d\ge0\). Squaring gives

\[2d+3=d^2,\]

so \((d-3)(d+1)=0\). The candidates are 3 and \(-1\), but only \(d=3\) satisfies the domain and the original equation.

Try It. Solve \(\sqrt{3h+4}=h\).
Check Your Work Squaring gives \(h^2-3h-4=0\), so \(h=4\) or \(-1\). Only \(h=4\) checks.

Applied Problem: Finding flow from detention time. A contact tank has volume 540 m³ and must provide a detention time of 45 min. Rearrange \(t=V/Q\) to solve for \(Q\), then calculate the required flow in cubic metres per minute and litres per second.

Check Your Work \(Q=V/t=540/45=12\) m³/min. Converting gives \(12{,}000\) L/min, then \(12{,}000/60=200\) L/s.

1.13.2 Practice Problems

  1. Solve \(d=vt\) for \(t\).
    Check Your Work \(t=d/v\), \(v\ne0\).
  2. Solve \(V=LWH\) for \(H\).
    Check Your Work \(H=V/(LW)\), \(LW\ne0\).
  3. Solve \(C=5(F-32)/9\) for \(F\).
    Check Your Work \(F=9C/5+32\).
  4. Solve \(\sqrt{x+1}=3\).
    Check Your Work \(x=8\).
  5. Solve \(\sqrt{2x-1}=x-2\).
    Check Your Work The candidates are 1 and 5; only \(x=5\) checks.
  6. Rearrange the loading-rate formula \(R=Q/A\) to solve for required filter area \(A\).
    Check Your Work \(RA=Q\), so \(A=Q/R\), with \(R\ne0\).
  7. A tank has \(V=\pi r^2h\). Solve for \(h\), then find \(h\) when \(V=100\) m³ and \(r=2.5\) m.
    Check Your Work \(h=V/(\pi r^2)\). Thus \(h=100/[\pi(2.5)^2]\approx5.09\) m.
  8. Solve the water-depth equation \(\sqrt{2d+8}=d\).
    Check Your Work Squaring gives \(d^2-2d-8=0\), so \(d=4\) or \(-2\). Only \(d=4\) checks.