Chapter 12.1, Problem 49WE

a.

To determine

To calculate: The number of real-number solutions for the equation $a{\left(x+b\right)}^2=c$ when $a>0\text{ and }c>0$ .

Answer to Problem 49WE

The number of real-number solutions are $2$ .

Explanation of Solution

Given information:

The equation $a{\left(x+b\right)}^2=c$ and conditions $a>0\text{ and }c>0$ .

Calculation:

Consider the provided equation $a{\left(x+b\right)}^2=c$ and conditions $a>0\text{ and }c>0$ .

Recall that the quadratic equation $y^2=m$ has 2 real number roots if $m>0$ that are $y=\pm m$ .

Rewrite the equation $a{\left(x+b\right)}^2=c$ as ${\left(x+b\right)}^2=\frac{c}{a}$ that is divide both sides by a .

Now, let $\frac{c}{a}=m$ . Since, it is provided $a>0\text{ and }c>0$ therefore, $m>0$ as division of two positive integers is positive.

And according to condition the equation has 2 real number roots.

  $\begin{array}{c}a{\left(x+b\right)}^2=c\\ {\left(x+b\right)}^2=\frac{c}{a}\\ x+b=\pm \sqrt{\frac{c}{a}}\\ x=\pm \sqrt{\frac{c}{a}}-b\end{array}$

And the roots are $x=\pm \sqrt{\frac{c}{a}}-b$

Thus, the number of real-number solutions are $2$ .

b.

To determine

To calculate: The number of real-number solutions for the equation $a{\left(x+b\right)}^2=c$ when $a<0\text{ and }c<0$ .

Answer to Problem 49WE

The number of real-number solutions are $2$ .

Explanation of Solution

Given information:

The equation $a{\left(x+b\right)}^2=c$ and conditions $a<0\text{ and }c<0$ .

Calculation:

Consider the provided equation $a{\left(x+b\right)}^2=c$ and conditions $a<0\text{ and }c<0$ .

Recall that the quadratic equation $y^2=m$ has 2 real number roots if $m>0$ that are $y=\pm m$ .

Rewrite the equation $a{\left(x+b\right)}^2=c$ as ${\left(x+b\right)}^2=\frac{c}{a}$ that is divide both sides by a .

Now, let $\frac{c}{a}=m$ . Since, it is provided $a<0\text{ and }c<0$ therefore, $m>0$ as negative sign of both will cancel out and positive value is obtained.

And according to condition the equation has 2 real number roots.

  $\begin{array}{c}-a{\left(x+b\right)}^2=-c\\ {\left(x+b\right)}^2=\frac{-c}{-a}\\ x+b=\pm \sqrt{\frac{c}{a}}\\ x=\pm \sqrt{\frac{c}{a}}-b\end{array}$

And the roots are $x=\pm \sqrt{\frac{c}{a}}-b$ .

Thus, the number of real-number solutions are $2$ .

c.

To determine

To calculate: The number of real-number solutions for the equation $a{\left(x+b\right)}^2=c$ when $a<0\text{ and }c>0$ .

Answer to Problem 49WE

The number of real-number solutions is $0$ .

Explanation of Solution

Given information:

The equation $a{\left(x+b\right)}^2=c$ and conditions $a<0\text{ and }c>0$ .

Calculation:

Consider the provided equation $a{\left(x+b\right)}^2=c$ and conditions $a<0\text{ and }c>0$ .

Recall that the quadratic equation $y^2=m$ has no real number roots if $m<0$ .

Rewrite the equation $a{\left(x+b\right)}^2=c$ as ${\left(x+b\right)}^2=\frac{c}{a}$ that is divide both sides by a .

Now, let $\frac{c}{a}=m$ . Since, it is provided $a<0\text{ and }c>0$ therefore, $m<0$ as negative sign of will not cancel out and negative value is obtained.

And according to condition the equation has 0 real number roots.

Thus, the number of real-number solutions is 0.

d.

To determine

To calculate: The number of real-number solutions for the equation $a{\left(x+b\right)}^2=c$ when $a>0\text{ and }c=0$ .

Answer to Problem 49WE

The number of real-number solutions is $1$ .

Explanation of Solution

Given information:

The equation $a{\left(x+b\right)}^2=c$ and conditions $a>0\text{ and }c=0$ .

Calculation:

Consider the provided equation $a{\left(x+b\right)}^2=c$ and conditions $a>0\text{ and }c=0$ .

Recall that the quadratic equation $y^2=m$ has 1 real number roots if $m=0$ that are $y=0$ .

Rewrite the equation $a{\left(x+b\right)}^2=c$ as ${\left(x+b\right)}^2=\frac{c}{a}$ that is divide both sides by a .

Now, let $\frac{c}{a}=m$ . Since, it is provided $a>0\text{ and }c=0$ therefore, $m=0$ as 0 divided by any number is 0 only.

And according to condition the equation has 1 real number roots.

  $\begin{array}{c}{\left(x+b\right)}^2=0\\ x+b=0\\ x=-b\end{array}$

And the root is $x=-b$ .

Thus, the number of real-number solutions is $1$ .