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)^{(1)/(4)}=2`

- You need to apply this property:

`a^{(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: