Chapter 9.3, Problem 49PFA

a.

To determine

To identify: The quadratic equation which have solutions of $-2\mathrm{and}5$ .

Answer to Problem 49PFA

The option $\mathrm{A}$ has solution $-2\mathrm{and}5$ .

Explanation of Solution

Given information: The given equation is $x^2-3x-10=0$ .

Calculation: : The given equation is $x^2-3x-10=0$ .

The given zeroes are $-2\mathrm{and}5$ .

Sum of zeroes $=-2+5=3$ .

Product of zeroes $=-2\times 5=-10$

Expression of equation is $x^2-(sum)x+\mathrm{product}=0$

Then equation $=x^2-3x-10=0$

This equation is same as equation $\mathrm{A}$ .

Thus, the option $\mathrm{A}$ has solution $-2\mathrm{and}5$ .

b.

To determine

To identify: the quadratic equation which have solutions of $-2\mathrm{and}5$ .

Answer to Problem 49PFA

The option $\mathrm{B}$ has not solution $-2\mathrm{and}5$ .

Explanation of Solution

Given information: The given equation is $x^2+3x=10$ .

Calculation: : The given equation is $x^2+3x=10$ .

The given zeroes are $-2\mathrm{and}5$ .

Sum of zeroes $=-2+5=3$ .

Product of zeroes $=-2\times 5=-10$

Expression of equation is $x^2-(sum)x+\mathrm{product}=0$

Then equation $=x^2-3x-10=0$

Retype the equation:

  $x^2-3x=10$ this equation is not same as the equation $\mathrm{B}$ .

Thus, the option $\mathrm{B}$ has not solution $-2\mathrm{and}5$ .

c.

To determine

To identify: the quadratic equation which have solutions of $-2\mathrm{and}5$ .

Answer to Problem 49PFA

The option $\mathrm{C}$ has solution $-2\mathrm{and}5$ .

Explanation of Solution

Given information: The given equation is $2x^2-6x=20$ .

Calculation: : The given equation is $2x^2-6x=20$ .

The given zeroes are $-2\mathrm{and}5$ .

Sum of zeroes $=-2+5=3$ .

Product of zeroes $=-2\times 5=-10$

Expression of equation is $x^2-(sum)x+\mathrm{product}=0$

Then equation $=x^2-3x-10=0$

Retype the equation:

  $\begin{array}{l}{x}^2-3x-10=0\\ \mathrm{multiply}\mathrm{by}2\mathrm{both}\mathrm{sides}\\ 2x^2-6x-20=0\times 2\\ 2x^2-6x-20=0\end{array}$

This equation is same as equation $\mathrm{C}$ .

Thus, the option $\mathrm{C}$ has solution $-2\mathrm{and}5$ .

d.

To determine

To identify: the quadratic equation which have solutions of $-2\mathrm{and}5$ .

Answer to Problem 49PFA

The option $\mathrm{D}$ has not solution $-2\mathrm{and}5$ .

Explanation of Solution

Given information: The given equation is $-x^2+10=3x$ .

Calculation: The given equation is $-x^2+10=3x$ .

The given zeroes are $-2\mathrm{and}5$ .

Sum of zeroes $=-2+5=3$ .

Product of zeroes $=-2\times 5=-10$

Expression of equation is $x^2-(sum)x+\mathrm{product}=0$

Then equation $=x^2-3x-10=0$

Retype the equation:

  $\begin{array}{l}{x}^2-3x-10=0\\ \mathrm{multiply}\mathrm{by}-1\mathrm{both}\mathrm{sides}\\ -x^2+3x+10=0\times (-1)\\ -x^2+3x+10=0\\ -x^2+10=-3x\end{array}$

This equation is not same as the equation $\mathrm{D}$ .

Thus, the option $\mathrm{D}$ has not solution $-2\mathrm{and}5$ .

e.

To determine

To identify: the quadratic equation which have solutions of $-2\mathrm{and}5$ .

Answer to Problem 49PFA

The option $\mathrm{E}$ has solution $-2\mathrm{and}5$ .

Explanation of Solution

Given information: The given equation is $-x^2+3x+10=0$ .

Calculation: The given equation is $-x^2+3x+10=0$ .

The given zeroes are $-2\mathrm{and}5$ .

Sum of zeroes $=-2+5=3$ .

Product of zeroes $=-2\times 5=-10$

Expression of equation is $x^2-(sum)x+\mathrm{product}=0$

Then equation $=x^2-3x-10=0$

Retype the equation:

  $\begin{array}{l}{x}^2-3x-10=0\\ \mathrm{multiply}\mathrm{by}-1\mathrm{both}\mathrm{sides}\\ -x^2+3x+10=0\times (-1)\\ -x^2+3x+10=0\end{array}$

This equation is same as equation $\mathrm{E}$ .

Thus, the option $\mathrm{E}$ has solution $-2\mathrm{and}5$ .