Sum and Product of Roots

The quadratic formula
 

$ x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} $

 

give the roots of a quadratic equation which may be real or imaginary. The ± sign in the radical indicates that
 

$ x_1 = \dfrac{-b + \sqrt{b^2-4ac}}{2a} $   and   $ x_2 = \dfrac{-b - \sqrt{b^2-4ac}}{2a} $

 

where x1 and x2 are the roots of the quadratic equation ax2 + bx + c = 0. The sum of roots x1 + x2 and the product of roots x1·x2 are common to problems involving quadratic equation.
 

Derivation of Quadratic Formula

The roots of a quadratic equation ax2 + bx + c = 0 is given by the quadratic formula
 

$ x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} $

 

The derivation of this formula can be outlined as follows:

  1. Divide both sides of the equation ax2 + bx + c = 0 by a.
  2. Transpose the quantity c/a to the right side of the equation.
  3. Complete the square by adding b2 / 4a2 to both sides of the equation.
  4. Factor the left side and combine the right side.
  5. Extract the square-root of both sides of the equation.
  6. Solve for x by transporting the quantity b / 2a to the right side of the equation.
  7. Combine the right side of the equation to get the quadratic formula.

See the derivation below.
 

Solution to Problem 250 Statically Indeterminate

Problem 250
In the assembly of the bronze tube and steel bolt shown in Fig. P-250, the pitch of the bolt thread is p = 1/32 in.; the cross-sectional area of the bronze tube is 1.5 in.2 and of steel bolt is 3/4 in.2 The nut is turned until there is a compressive stress of 4000 psi in the bronze tube. Find the stresses if the nut is given one additional turn. How many turns of the nut will reduce these stresses to zero? Use Ebr = 12 × 106 psi and Est = 29 × 106 psi.
 

Figure P-250

 

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