Chapter A.7, Problem 1MP

(A)

To determine

To solve: The equation $x^2-6=0$ using square root property.

Answer to Problem 1MP

The solutions of the equation $x^2-6=0$ are $x=\pm \sqrt{6}$.

Explanation of Solution

Theorem used:

Square Root Property:

If $a^2=b$, then $a=\pm \sqrt{b}$.

Calculation:

The given equation is $x^2-6=0$.

Obtain the solution of the equation $x^2-6=0$ as follows.

$\begin{array}{l}{x}^2-6=0\\ x^2=6\\ x=\pm \sqrt{6}\left[\because \text{Square root property}\right]\end{array}$

Thus, the solutions of the equation $x^2-6=0$ are $x=\pm \sqrt{6}$.

(B)

To determine

To solve: The equation $3x^2-12=0$ using square root property.

Answer to Problem 1MP

The solutions of the equation $3x^2-12=0$ are $x=\pm 2$.

Explanation of Solution

Obtain the solution of the equation $3x^2-12=0$ as follows.

$\begin{array}{l}3x^2-12=0\\ 3x^2=12\\ x^2=4\\ x=\pm 2\left[\because \text{Square root property}\right]\end{array}$

Thus, the solutions of the equation $3x^2-12=0$ are $x=\pm 2$.

(C)

To determine

To solve: The equation $x^2+4=0$ using square root property.

Answer to Problem 1MP

The equation $x^2+4=0$ has no real solution.

Explanation of Solution

Obtain the solution of the equation $x^2+4=0$.

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

Since, there exist no real number whose square is negative, the equation $x^2+4=0$ has no real roots.

Thus, the equation $x^2+4=0$ has no real solution.

(D)

To determine

To solve: The equation ${\left(x+5\right)}^2=1$ using square root property.

Answer to Problem 1MP

The solutions of the equations ${\left(x+5\right)}^2=1$ are $x=-4,-6$.

Explanation of Solution

Obtain the solution of the equation ${\left(x+5\right)}^2=1$ using square-root property.

$\begin{array}{l}{\left(x+5\right)}^2=1\\ x+5=\pm \sqrt{1}\left[\because \text{Square root property}\right]\\ x=-5\pm 1\\ x=-5+1\text{or}-5-1\end{array}$

That is, $x=-4\text{or}-6$.

Thus, the solutions of the equations ${\left(x+5\right)}^2=1$ are $x=-4,-6$.