1.12 Solving linear and quadratic equations

A linear equation has the form \(ax+b=0\), with \(a\ne0\). A quadratic has the form \(ax^2+bx+c=0\), with \(a\ne0\). Quadratics may be solved by factoring or by the quadratic formula.

Quadratic formula The solution to \(ax^2+bx+c=0\) with \(a\ne0\). is given by the quadratic formula

\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.\]

The discriminant \(D=b^2-4ac\) indicates two real solutions when \(D>0\), one repeated real solution when \(D=0\), and no real solutions when \(D<0\). Evaluate it before taking the square root. When substituting negative coefficients into the formula, use parentheses.

1.12.1 Solving by factoring

Factoring is efficient after zero has been placed on one side. The zero-product property then gives one equation from each factor.

Zero-product property If \(A\) and \(B\) are real numbers, then \(AB = 0\) if and only if \(A = 0\) or \(B = 0\).

The equation \(3x^2-10x=8\) becomes \((3x+2)(x-4)=0\), giving \(x=-2/3\) or \(x=4\). Both check algebraically. An application adds a second check: a negative dimension or time is rejected, while a negative signed elevation or rate may be meaningful under a stated convention.

Worked Example: Finding basin dimensions

A basin has area 96 m² and length 4 m greater than width \(w\). Then

\[w(w+4)=96,\]

\[w^2+4w-96=(w+12)(w-8)=0.\]

The algebraic candidates are \(-12\) and 8. Rejecting the negative dimension gives width 8 m and length 12 m.

Try It. A rectangular tank has area 60 m² and is 7 m longer than it is wide. Find its dimensions.
Check Your Work \(w(w+7)=60\), so \((w+12)(w-5)=0\). The tank is 5 m by 12 m.

Factoring is efficient only after the equation is written with zero on one side. The zero-product property then states that \(AB=0\) exactly when \(A=0\) or \(B=0\). An equation such as \((x+2)^2=25\) can instead use the square-root property, giving \(x+2=\pm5\). In applications, check every algebraic solution against dimensions, signs, and other physical restrictions.

Worked Example: Solving a squared depth relationship

If \((d+1)^2=16\), then the square-root property gives

\[d+1=\pm4.\]

Thus \(d=3\) or \(d=-5\). If \(d\) represents water depth, only \(d=3\) m is physically meaningful.

Try It. Solve \((h-2)^2=25\) and identify the physically meaningful positive depth.
Check Your Work \(h-2=\pm5\), so \(h=7\) or \(h=-3\). The meaningful depth is 7 m.

1.12.2 Using the quadratic formula

The quadratic formula is especially useful when a quadratic does not factor conveniently over the integers. First identify \(a\), \(b\), and \(c\) from standard form \(ax^2+bx+c=0\). Substitute signed coefficients with parentheses, simplify the discriminant, and divide the entire numerator by \(2a\).

Worked Example: Applying the quadratic formula

For \(2x^2-3x-1=0\),

\[x=\frac{3\pm\sqrt{(-3)^2-4(2)(-1)}}{4}=\frac{3\pm\sqrt{17}}4.\]

Try It. Solve \(x^2-4x-1=0\).
Check Your Work \(x=2\pm\sqrt5\).

Worked Example: Solving a nonfactorable area equation

Suppose a design equation is \(3d^2+2d-7=0\). The quadratic formula gives

\[d=\frac{-2\pm\sqrt{2^2-4(3)(-7)}}{2(3)}=\frac{-1\pm\sqrt{22}}3.\]

If \(d\) is a depth, retain only \(d=(-1+\sqrt{22})/3\approx1.23\).

Try It. Solve \(2x^2+x-4=0\) with the quadratic formula.
Check Your Work \(x=(-1\pm\sqrt{33})/4\).

1.12.3 Interpreting the discriminant

The discriminant is the expression \(D=b^2-4ac\) under the radical in the quadratic formula. Its sign determines the number of real solutions before the equation is solved: two when \(D>0\), one repeated solution when \(D=0\), and none when \(D<0\).

Worked Example: Interpreting the discriminant

For \(x^2+3x+2=0\), the discriminant is

\[D=3^2-4(1)(2)=1>0.\]

The equation has two distinct real solutions. Factoring confirms that they are \(-1\) and \(-2\).

Try It. Use the discriminant to classify the solutions of \(x^2-6x+9=0\).
Check Your Work \(D=36-36=0\), so there is one repeated real solution.

Applied Problem: Designing a rectangular contact basin. A basin must have surface area 180 m². Its length will be 7 m greater than its width. Form and solve a quadratic equation for the dimensions, rejecting any physically impossible result.

Check Your Work Let the width be \(w\). Then \(w(w+7)=180\), so \(w^2+7w-180=0=(w+15)(w-12)\). The candidates are \(-15\) and 12. Reject the negative value. The basin is 12 m wide and 19 m long.

1.12.4 Practice Problems

  1. Solve \(7x-9=26\).
    Check Your Work \(x=5\).
  2. Solve \(x^2-9=0\).
    Check Your Work \(x=\pm3\).
  3. Solve \(x^2+7x+12=0\).
    Check Your Work \(x=-3,-4\).
  4. Determine the number of real solutions of \(3x^2+2x+5=0\).
    Check Your Work No real solutions because \(D=-56\).
  5. A square tank floor has area 49 m². Find its positive side length.
    Check Your Work 7 m.
  6. A rectangular wet well has area 88 m² and is 3 m longer than it is wide. Find its dimensions.
    Check Your Work \(w(w+3)=88\), so \((w+11)(w-8)=0\). The dimensions are 8 m by 11 m.
  7. A design equation is \(2d^2-5d-3=0\). Find the physically meaningful positive solution.
    Check Your Work \((2d+1)(d-3)=0\), so \(d=-1/2\) or 3. The positive solution is 3.
  8. Use the discriminant to determine how many real operating points satisfy \(4q^2-4q+5=0\).
    Check Your Work \(D=(-4)^2-4(4)(5)=-64<0\), so there are no real operating points.