Math / Algebra

QuestionSolve the equation by factoring. Check your solution. If there are multiple solutions, list the solutions from leas greatest separated by a comma. $x^{2}+4=0$ $\square$

Studdy Solution

STEP 1

What is this asking? We need to find the values of $x$ that make the equation $x^2 + 4 = 0$ true, and then double-check our answers! Watch out! Remember, even though we're factoring, we might encounter solutions that involve imaginary numbers.
Don't let that scare you!

STEP 2

1. Isolate the $x^2$ term
2. Take the square root of both sides
3. Check the solutions

STEP 3

Alright, let's isolate that $x^2$ term!
We've got $x^2 + 4 = 0$ .
To get $x^2$ by itself, we need to move that 4 to the other side of the equation.
We can do this by adding -4 to both sides: $x^2 + 4 + (-4) = 0 + (-4)$ $x^2 = -4$ Now our equation looks much simpler!

STEP 4

Now, let's apply the square root to both sides of the equation $x^2 = -4$ to get $x$ all alone.
Remember, when we take the square root of both sides, we need to consider both the positive and negative roots: $\sqrt{x^2} = \pm\sqrt{-4}$ $\sqrt{x^2} = \pm\sqrt{4 \cdot -1}$ $\sqrt{x^2} = \pm\sqrt{4} \cdot \sqrt{-1}$ Since $\sqrt{-1}$ is defined as $i$ , we get: $x = \pm 2i$ So, we have two possible solutions: $x = 2i$ and $x = -2i$ .

STEP 5

Let's check our first solution, $x = 2i$ , by plugging it back into the original equation: $(2i)^2 + 4 = 0$ $4i^2 + 4 = 0$ Since $i^2 = -1$ : $4(-1) + 4 = 0$ $-4 + 4 = 0$ $0 = 0$ Woohoo! It works!

STEP 6

Now, let's check our second solution, $x = -2i$ : $(-2i)^2 + 4 = 0$ $4i^2 + 4 = 0$ Again, since $i^2 = -1$ : $4(-1) + 4 = 0$ $-4 + 4 = 0$ $0 = 0$ Fantastic! This one works too!

STEP 7

Our solutions are $x = -2i$ and $x = 2i$ .
Listing them from least to greatest, we get $-2i, 2i$ .

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QuestionSolve the equation by factoring. Check your solution. If there are multiple solutions, list the solutions from leas greatest separated by a comma. $x^{2}+4=0$ $\square$

Studdy Solution

STEP 1

What is this asking? We need to find the values of $x$ that make the equation $x^2 + 4 = 0$ true, and then double-check our answers! Watch out! Remember, even though we're factoring, we might encounter solutions that involve imaginary numbers.
Don't let that scare you!

STEP 2

1. Isolate the $x^2$ term
2. Take the square root of both sides
3. Check the solutions

STEP 3

Alright, let's isolate that $x^2$ term!
We've got $x^2 + 4 = 0$ .
To get $x^2$ by itself, we need to move that 4 to the other side of the equation.
We can do this by adding -4 to both sides: $x^2 + 4 + (-4) = 0 + (-4)$ $x^2 = -4$ Now our equation looks much simpler!

STEP 4

STEP 5

Let's check our first solution, $x = 2i$ , by plugging it back into the original equation: $(2i)^2 + 4 = 0$ $4i^2 + 4 = 0$ Since $i^2 = -1$ : $4(-1) + 4 = 0$ $-4 + 4 = 0$ $0 = 0$ Woohoo! It works!

STEP 6

Now, let's check our second solution, $x = -2i$ : $(-2i)^2 + 4 = 0$ $4i^2 + 4 = 0$ Again, since $i^2 = -1$ : $4(-1) + 4 = 0$ $-4 + 4 = 0$ $0 = 0$ Fantastic! This one works too!

STEP 7

Our solutions are $x = -2i$ and $x = 2i$ .
Listing them from least to greatest, we get $-2i, 2i$ .

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QuestionSolve the equation by factoring. Check your solution. If there are multiple solutions, list the solutions from leas greatest separated by a comma. x2+4=0x^{2}+4=0x2+4=0 □\square□

Studdy Solution

STEP 1

What is this asking? We need to find the values of xxx that make the equation x2+4=0x^2 + 4 = 0x2+4=0 true, and then double-check our answers! Watch out! Remember, even though we're factoring, we might encounter solutions that involve imaginary numbers.Don't let that scare you!

STEP 2

1. Isolate the x2x^2x2 term2. Take the square root of both sides3. Check the solutions

STEP 3

STEP 4

STEP 5

Let's **check** our first solution, x=2ix = 2ix=2i, by plugging it back into the **original equation**: (2i)2+4=0(2i)^2 + 4 = 0(2i)2+4=0 4i2+4=04i^2 + 4 = 04i2+4=0Since i2=−1i^2 = -1i2=−1: 4(−1)+4=04(-1) + 4 = 04(−1)+4=0 −4+4=0-4 + 4 = 0−4+4=00=00 = 00=0Woohoo! It works!

STEP 6

Now, let's **check** our second solution, x=−2ix = -2ix=−2i: (−2i)2+4=0(-2i)^2 + 4 = 0(−2i)2+4=0 4i2+4=04i^2 + 4 = 04i2+4=0Again, since i2=−1i^2 = -1i2=−1: 4(−1)+4=04(-1) + 4 = 04(−1)+4=0 −4+4=0-4 + 4 = 0−4+4=00=00 = 00=0Fantastic! This one works too!

STEP 7

Our solutions are x=−2ix = -2ix=−2i and x=2ix = 2ix=2i.Listing them from least to greatest, we get −2i,2i-2i, 2i−2i,2i.

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QuestionSolve the equation by factoring. Check your solution. If there are multiple solutions, list the solutions from leas greatest separated by a comma. $x^{2}+4=0$ $\square$

What is this asking? We need to find the values of $x$ that make the equation $x^2 + 4 = 0$ true, and then double-check our answers! Watch out! Remember, even though we're factoring, we might encounter solutions that involve imaginary numbers.
Don't let that scare you!

1. Isolate the $x^2$ term
2. Take the square root of both sides
3. Check the solutions

Let's check our first solution, $x = 2i$ , by plugging it back into the original equation: $(2i)^2 + 4 = 0$ $4i^2 + 4 = 0$ Since $i^2 = -1$ : $4(-1) + 4 = 0$ $-4 + 4 = 0$ $0 = 0$ Woohoo! It works!

Now, let's check our second solution, $x = -2i$ : $(-2i)^2 + 4 = 0$ $4i^2 + 4 = 0$ Again, since $i^2 = -1$ : $4(-1) + 4 = 0$ $-4 + 4 = 0$ $0 = 0$ Fantastic! This one works too!

Our solutions are $x = -2i$ and $x = 2i$ .
Listing them from least to greatest, we get $-2i, 2i$ .