Chapter ISG, Problem 9.6.4P

To determine

To solve equation.

Answer to Problem 9.6.4P

  $x_1=\frac{-b+\sqrt{b^2-4c}}{2}$

  $x_2=\frac{-b-\sqrt{b^2-4c}}{2}$

Statement: if b and c are negative, there will always be at least one real solution.

Statement: if b and c are negative, there will sometimes be at least one real solution.

Explanation of Solution

Given information:

  $x^2+bx+c=0$

Formula used:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

  $x^2+bx+c=0$

  $a=1$

  $b=b$

  $c=c$

Using formula

  $x=\frac{-b\pm \sqrt{b^2-4c}}{2}$

  $x_1=\frac{-b+\sqrt{b^2-4c}}{2}$

  $x_2=\frac{-b-\sqrt{b^2-4c}}{2}$

Statement: if b and c are negative, there will always be at least one real solution.

if b and c are negative

  $b^2-4ac>0$

Then always at least real solution

Statement: if b and c are negative, there will sometimes be at least one real solution.

if b and c are positive

  $b^2-4ac>0$ or $b^2-4ac<0$

Then sometimes complex nonreal solutions