Chapter 4.2, Problem 3CFU

(a)

To determine

To describe: The discriminant of the equation represented by the given graph.

Answer to Problem 3CFU

  $\Delta =b^2-4ac=0$

Explanation of Solution

Given information: The graph

  

Calculation:

Since the parabola from the figure, has the vertex tangent to x-axis, that means that the discriminant is zero. a, b, c are the coefficients of the quadratic equation.

  $\Delta =b^2-4ac=0$

The solutions of the quadratic $x_1\text{ and }{x}_2$ depend on the discriminant delta.

  $x_{1,2}=\left(-b\pm \sqrt{\Delta }\right)/2a$

Since the discriminant is zero, the quadratic has two equal solutions: $x_1=x_2$

  $\Delta =0=>x_{1,2}=\left(-b\pm 0\right)/2a=-b/2a$

  $\Delta =b^2-4ac=0$

(b)

To determine

To describe: The discriminant of the equation represented by the given graph.

Answer to Problem 3CFU

  $\Delta <0$

Explanation of Solution

Given information: The graph

  

Calculation:

Since the parabola is not intercepting x-axis, then the discriminant is negative.

  $\Delta =b^2-4ac<0$

Since the discriminant delta is negative, we cannot determine the square root of a negative number, therefore there are no real roots for the quadratic whose discriminant is negative. But there are complex roots.

  $\begin{array}{l}{x}_{1,2}=\left(-b\pm \sqrt{\Delta }\right)/2a=\text{impossible}\\ \Delta <0\end{array}$

(c)

To determine

To describe: The discriminant of the equation represented by the given graph.

Answer to Problem 3CFU

  $\Delta >0$

Explanation of Solution

Given information: The graph

  

Calculation:

Since the parabola is intercepting x-axis, then the discriminant is positive.

  $\Delta =b^2-4ac>0$

Since delta is positive, the quadratic will have two different real solutions $x_1\text{ and }{x}_2$

  $\begin{array}{l}\Delta >0=>x_1\ne x_2=\left(-b\pm \sqrt{\Delta }\right)/2a\\ \Delta >0\end{array}$