Chapter 1.5, Problem 36E

Solution

a.

To write: an equation of the form $x^2=s$ that has two real solutions.

  $x^2=1$ .

Property Used:

  $\left(a-b\right)\left(a+b\right)=a^2-b^2$

Zero Product property: If product of two expressions is zero, then either one or both the expressions equal zero.

Calculation:

Let $s=1$ .

Then, the equation becomes $x^2=1$ .

Consider, $x^2=1$ .

This is equivalent to,

  $\begin{array}{c}{x}^2-1=0\\ x^2-1^2=0\\ \left(x-1\right)\left(x+1\right)=0\end{array}$

Then, by zero product property, $x-1=0$ or $x+1=0$ .

Solve for x , $x=1$ or $x=-1$ .

Thus, $x^2=1$ has two real solutions 1 and $-1$ .

b.

To write: an equation of the form $x^2=s$ that has exactly one real solution.

  $x^2=0$

Property Used:

  $\left(a-b\right)\left(a+b\right)=a^2-b^2$

Zero Product property: If product of two expressions is zero, then either one or both the expressions equal zero.

Calculation:

Let $s=0$

Then, the equation becomes $x^2=0$

Consider,

  $x^2=0$

Take square root of both sides,

  $\begin{array}{c}\sqrt{x^2}=\pm \sqrt{0}\\ x=\pm 0\end{array}$

But it is known that $\pm 0=0$

Thus, $x^2=0$ has exactly one real solution, that is 0.

c.

To write: an equation of the form $x^2=s$ that has no real solutions.

  $x^2=-1$

Property Used:

  $i=\sqrt{-1}$

Zero Product property: If product of two expressions is zero, then either one or both the expressions equal zero.

Calculation:

Let $s=-1$

Then, the equation becomes $x^2=-1$

Consider,

  $x^2=-1$

Take square root of both sides,

  $\begin{array}{c}\sqrt{x^2}=\pm \sqrt{-1}\\ x=\pm i\end{array}$

Thus, $x^2=-1$ has no real solutions but two complex solutions $i$ and $-i$ .