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 $-2\mathrm{and}5$.