Math / Algebra

QuestionFind all real solutions of the equation $(x-2)^{2}=16$ . $x_{1}=\square \text { and } x_{2}=\square \text { with } x_{1}<x_{2}$

Studdy Solution

STEP 1

What is this asking? We need to find two real numbers, $x_1$ and $x_2$ , that solve the equation $(x-2)^2 = 16$ , where $x_1$ is smaller than $x_2$ . Watch out! Remember that squared equations can have two solutions, one positive and one negative!
Don't lose one along the way.

STEP 2

1. Isolate the squared term
2. Take the square root of both sides
3. Solve for x for both positive and negative roots
4. Order solutions

STEP 3

The equation is already in a nice form, with the squared term $(x-2)^2$ isolated on one side.
So, we're good to go!

STEP 4

Let's take the square root of both sides of the equation $(x-2)^2 = 16$ .
Remember, when we take the square root, we consider both the positive and negative roots.
This gives us: $\sqrt{(x-2)^2} = \pm\sqrt{16}$

STEP 5

Simplifying, we get: $x-2 = \pm 4$ This is because the square root of $(x-2)^2$ is $(x-2)$ and the square root of 16 is 4.

STEP 6

First, let's consider the positive root: $x-2 = 4$ To isolate $x$ , we add 2 to both sides of the equation: $x-2+2 = 4+2$ $x = 6$

STEP 7

Now, let's consider the negative root: $x-2 = -4$ Again, we add 2 to both sides: $x-2+2 = -4+2$ $x = -2$

STEP 8

We found two solutions: $x=6$ and $x=-2$ .
Since $-2$ is less than $6$ , we have $x_1 = -2$ and $x_2 = 6$ .

STEP 9

$x_1 = -2$ and $x_2 = 6$ .

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QuestionFind all real solutions of the equation $(x-2)^{2}=16$ . $x_{1}=\square \text { and } x_{2}=\square \text { with } x_{1}<x_{2}$

Studdy Solution

STEP 1

What is this asking? We need to find two real numbers, $x_1$ and $x_2$ , that solve the equation $(x-2)^2 = 16$ , where $x_1$ is smaller than $x_2$ . Watch out! Remember that squared equations can have two solutions, one positive and one negative!
Don't lose one along the way.

STEP 2

1. Isolate the squared term
2. Take the square root of both sides
3. Solve for x for both positive and negative roots
4. Order solutions

STEP 3

The equation is already in a nice form, with the squared term $(x-2)^2$ isolated on one side.
So, we're good to go!

STEP 4

Let's take the square root of both sides of the equation $(x-2)^2 = 16$ .
Remember, when we take the square root, we consider both the positive and negative roots.
This gives us: $\sqrt{(x-2)^2} = \pm\sqrt{16}$

STEP 5

Simplifying, we get: $x-2 = \pm 4$ This is because the square root of $(x-2)^2$ is $(x-2)$ and the square root of 16 is 4.

STEP 6

First, let's consider the positive root: $x-2 = 4$ To isolate $x$ , we add 2 to both sides of the equation: $x-2+2 = 4+2$ $x = 6$

STEP 7

Now, let's consider the negative root: $x-2 = -4$ Again, we add 2 to both sides: $x-2+2 = -4+2$ $x = -2$

STEP 8

We found two solutions: $x=6$ and $x=-2$ .
Since $-2$ is less than $6$ , we have $x_1 = -2$ and $x_2 = 6$ .

STEP 9

$x_1 = -2$ and $x_2 = 6$ .

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QuestionFind all real solutions of the equation (x−2)2=16(x-2)^{2}=16(x−2)2=16. x1=□ and x2=□ with x1<x2x_{1}=\square \text { and } x_{2}=\square \text { with } x_{1}<x_{2}x1​=□ and x2​=□ with x1​<x2​

Studdy Solution

STEP 1

STEP 2

1. Isolate the squared term2. Take the square root of both sides3. Solve for x for both positive and negative roots4. Order solutions

STEP 3

The equation is already in a nice form, with the squared term (x−2)2(x-2)^2(x−2)2 isolated on one side.So, we're good to go!

STEP 4

Let's take the square root of both sides of the equation (x−2)2=16(x-2)^2 = 16(x−2)2=16.Remember, when we take the square root, we consider both the positive and negative roots.This gives us: (x−2)2=±16 \sqrt{(x-2)^2} = \pm\sqrt{16} (x−2)2​=±16​

STEP 5

Simplifying, we get: x−2=±4 x-2 = \pm 4 x−2=±4 This is because the square root of (x−2)2(x-2)^2(x−2)2 is (x−2)(x-2)(x−2) and the square root of **16** is **4**.

STEP 6

First, let's consider the positive root: x−2=4 x-2 = 4 x−2=4 To isolate xxx, we **add 2** to both sides of the equation: x−2+2=4+2 x-2+2 = 4+2 x−2+2=4+2 x=6 x = 6 x=6

STEP 7

Now, let's consider the negative root: x−2=−4 x-2 = -4 x−2=−4 Again, we **add 2** to both sides: x−2+2=−4+2 x-2+2 = -4+2 x−2+2=−4+2 x=−2 x = -2 x=−2

STEP 8

We found two solutions: x=6x=6x=6 and x=−2x=-2x=−2.Since −2-2−2 is less than 666, we have x1=−2x_1 = -2x1​=−2 and x2=6x_2 = 6x2​=6.

STEP 9

x1=−2x_1 = -2x1​=−2 and x2=6x_2 = 6x2​=6.

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QuestionFind all real solutions of the equation $(x-2)^{2}=16$ . $x_{1}=\square \text { and } x_{2}=\square \text { with } x_{1}<x_{2}$

1. Isolate the squared term
2. Take the square root of both sides
3. Solve for x for both positive and negative roots
4. Order solutions

The equation is already in a nice form, with the squared term $(x-2)^2$ isolated on one side.
So, we're good to go!

Let's take the square root of both sides of the equation $(x-2)^2 = 16$ .
Remember, when we take the square root, we consider both the positive and negative roots.
This gives us: $\sqrt{(x-2)^2} = \pm\sqrt{16}$

Simplifying, we get: $x-2 = \pm 4$ This is because the square root of $(x-2)^2$ is $(x-2)$ and the square root of 16 is 4.

First, let's consider the positive root: $x-2 = 4$ To isolate $x$ , we add 2 to both sides of the equation: $x-2+2 = 4+2$ $x = 6$

Now, let's consider the negative root: $x-2 = -4$ Again, we add 2 to both sides: $x-2+2 = -4+2$ $x = -2$

We found two solutions: $x=6$ and $x=-2$ .
Since $-2$ is less than $6$ , we have $x_1 = -2$ and $x_2 = 6$ .

$x_1 = -2$ and $x_2 = 6$ .