Chapter A.7, Problem 1MP

Use the square-root property to solve each equation.

$\begin{array}{cc}\text{(A) }{x}^2-6=0& \text{(B) }3x^2-12=0\\ \text{(C) }{x}^2+4=0& \text{(D) }{\left(x+5\right)}^2=1\end{array}$

To determine

To calculate : The solution for given equations by applying the square root property.

Answer to Problem 1MP

(A) The solution for $x$ is $\pm \sqrt{6}$.

( B ) The solution for $x$ is $\pm 2$.

( C ) There is no real solution for $x$.

( D ) The solution for $x$ is $-6$ or $-4$.

Explanation of Solution

Calculation :

(A)

Consider the equation, $x^2-6=0$.

Use the square root property to compute the value of $x$ as follows.

$\begin{array}{c}{x}^2-6=0\\ x^2=6\\ x=\pm \sqrt{6}\end{array}$

Thus, the solution for $x$ is $\pm \sqrt{6}$.

Now check the solutions in original problem:

Substitute $x=\sqrt{6}$ in equation $x^2-6=0$.

$\begin{array}{r}{\left(\sqrt{6}\right)}^2-6\stackrel{?}{=}0\\ 6-6\stackrel{?}{=}0\\ 0=0\end{array}$

Which is true.

Substitute $x=-\sqrt{6}$ in equation $x^2-6=0$.

$\begin{array}{r}{\left(-\sqrt{6}\right)}^2-6\stackrel{?}{=}0\\ 6-6\stackrel{?}{=}0\\ 0=0\end{array}$

Which is true.

Thus, the solution for $x$ is $\pm \sqrt{6}$.

( B )

Consider the equation, $3x^2-12=0$.

Use the square root property to compute the value of $x$ as follows.

$\begin{array}{c}3x^2-12=0\\ 3x^2=12\\ x=\pm \sqrt{4}\\ x=\pm 2\end{array}$

Thus, the solution for $x$ is $\pm 2$.

Now check the solutions in original equation $3x^2-12=0$.

For $x=2$.

$\begin{array}{r}3{\left(2\right)}^2-12\stackrel{?}{=}0\\ 12-12\stackrel{?}{=}0\\ 0=0\end{array}$

For $x=-2$.

$\begin{array}{r}3{\left(-2\right)}^2-12\stackrel{?}{=}0\\ 12-12\stackrel{?}{=}0\\ 0=0\end{array}$

So, the solutions are true.

Thus, the solution for $x$ is $\pm 2$.

( C )

Consider the equation, $x^2+4=0$.

Use the square root property to compute the value of $x$ as follows.

$\begin{array}{c}{x}^2+4=0\\ x^2=-4\end{array}$

Square of any real number cannot be negative.

Thus, there is no real solution for $x$.

(D)

Consider the equation, ${\left(x+5\right)}^2=1$.

Use the square root property to compute the value of $x$ as follows.

$\begin{array}{c}{\left(x+5\right)}^2=1\\ x+5=\pm \sqrt{1}\\ x+5=\pm 1\end{array}$

Solve further as,

$\begin{array}{c}x+5=1\text{ or }x+5=-1\\ x=-4\text{ or }x=-6\end{array}$

Now check the solutions in original equation $3x^2-12=0$.

For $x=-4$.

$\begin{array}{r}{\left(-4+5\right)}^2\stackrel{?}{=}1\\ {\left(1\right)}^2\stackrel{?}{=}1\\ 1=1\end{array}$

For $x=-6$.

$\begin{array}{r}{\left(-6+5\right)}^2\stackrel{?}{=}1\\ {\left(-1\right)}^2\stackrel{?}{=}1\\ 1=1\end{array}$

So, the solutions are true.

Thus, the solution for $x$ is $-6$ or $-4$.