Chapter 9.2, Problem 44HP

(a)

To determine

Write a quadratic equation that has one double root.

Answer to Problem 44HP

The equation: $x^2+10x+25=0\text{or}{(x+5)}^2=0$ . There is a double root at $x=-5$

Explanation of Solution

Given:

Concept Used:

A double root occurs when the curves vertex is on the x − axis.

Consider $x^2+10x+25=0\text{or}{(x+5)}^2=0$ . There is a double root at $x=-5$

Calculation:

A double root occurs when the curves vertex is on the x − axis. Consider $x^2+10x+25=0\text{or}{(x+5)}^2=0$. There is a double root at $x=-5$

Thus, the graph $x^2+10x+25=0\text{or}{(x+5)}^2=0$ , There is a double root at $x=-5$

(b)

To determine

Write a quadratic equation that has one rational root and one integral root.

Answer to Problem 44HP

The equation is $2x^2-23x+45=0$ and the zeros are: $9\text{and}\frac{5}{2}$

Explanation of Solution

Given:

Concept Used:

To have one rational root and one integral root, find two binomials where one equals an integer and there other a rational number.

Consider $2x^2-23x+45=0\text{or}(2x-5)(x-9)=0$

Calculation:

Consider the equation: $2x^2-23x+45=0\text{or}(2x-5)(x-9)=0$

Solve for x using the zero product property.

  $\begin{array}{l}2x-5=0\text{and}x-9=0\\ 2x=5\text{and}x=9\\ x=\frac{5}{2}\text{and}x=9\end{array}$ The zeros are $9\text{and}\frac{5}{2}$

Thus, the equation is $2x^2-23x+45=0$ and the zeros are: $9\text{and}\frac{5}{2}$

(c)

To determine

Write a quadratic equation that has two additive opposites.

Answer to Problem 44HP

The equation: $x^2-16=0\text{or}(x+4)(x-4)=0$ ; The zeros are $4\text{and}-4$

Explanation of Solution

Given:

Concept Used:

To have two distinct integral roots those are additive opposites.

Consider the equation: $x^2-16=0\text{or}(x+4)(x-4)=0$ ; The zeros are $4\text{and}-4$

Calculation:

Consider the equation: $x^2-16=0\text{or}(x+4)(x-4)=0$ ;

Solve for x using the zero product property.

  $\begin{array}{l}x+4=0\text{and}x-4=0\\ x=-4\text{and}x=4\end{array}$ The zeros are $-4\text{and}4$

Thus, the equation: $x^2-16=0\text{or}(x+4)(x-4)=0$ ; The zeros are $4\text{and}-4$