Question

4 Drag the tiles to the correct boxes. Not all tiles will be used. Match each equation with a value of x that satisfies it. 18 1 2 -3 5 CO ✓02 + 7 4 ♡ (ir — 2)4 Reset Next

Answer 1

You need to solve for "x" in each equation given in the exercise:

Equation 1

$\sqrt[]{x^2+7}=4$

- You need to apply the following property:

$(\sqrt[n]{a})^n=a$

Then:

$\begin{gathered} (\sqrt[]{x^2+7})^2=(4)^2 \\ \\ x^2+7=16 \end{gathered}$

- Subtract 7 from both sides of the equation:

$\begin{gathered} x^2+7-(7)=16-(7) \\ x^2=9 \end{gathered}$

- Finally, you have to take the square root of both sides of the equation:

$\begin{gathered} \sqrt[]{x^2}=\pm\sqrt[]{9} \\ \\ x_1=\sqrt[]{9}=3 \\ \\ x_2=-\sqrt[]{9}=-3 \end{gathered}$

In order to know which one satisfies the equation, let's substitute both values into the original equation and evaluate.

Then, for:

$x=3$

You get:

$\begin{gathered} \sqrt[]{(3)^2+7}=4 \\ \sqrt[]{16}=4 \\ 4=4\text{ (True)} \end{gathered}$

For:

$x=-3$

You get:

$\begin{gathered} \sqrt[]{(-3)^2+7}=4 \\ 4=4(\text{True)} \end{gathered}$

When a negative value has an even exponent, the result is always positive.

Equation 2

$(x-2)^{\frac{1}{4}}=2$

- You need to apply this property:

$a^{\frac{1}{n}}=\sqrt[n]{a}$

Then:

$\sqrt[4]{x-2}^{}=2$

- Now, applying the property:

$(\sqrt[n]{a})^n=a$

You get:

$\begin{gathered} (\sqrt[4]{x-2}^{})^4=(2)^4 \\ x-2=16 \end{gathered}$

- Finally, adding 2 to both sides of the equation, you get:

$\begin{gathered} x-2+(2)=16+(2) \\ x=18 \end{gathered}$

Check the answer:

$\begin{gathered} \sqrt[4]{18-2}^{}=2\text{ } \\ 2=2(\text{True)} \end{gathered}$

Equation 3

$\sqrt[3]{1-x}=-1$

- Knowing the properties explained before, you get:

$\begin{gathered} (\sqrt[3]{1-x})^3=(-1)^3 \\ 1-x=-1 \\ \end{gathered}$

- Now you can subtract 1 from both sides of the equation:

$\begin{gathered} 1-x-(1)=-1-(1) \\ -x=-2 \end{gathered}$

- Finally, you have to multiply both sides of the equation by -1:

$\begin{gathered} (-1)(-x)=(-2)(-1) \\ x=2 \end{gathered}$

Check the solution:

$\begin{gathered} \sqrt[3]{1-2}=-1 \\ \sqrt[3]{-1}=-1 \\ -1=-1(\text{True)} \end{gathered}$

The answer is: