Math / Algebra

QuestionSolve using the quadratic formula. $k^{2}-k-4=0$
Write your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth. $k=$ $\square$ or $k=$ $\square$ Submit

Studdy Solution

STEP 1

What is this asking? We need to find the values of $k$ that make the equation $k^2 - k - 4 = 0$ true, using the quadratic formula! Watch out! Remember the quadratic formula has a plus-or-minus sign, meaning we'll likely have two answers!
Also, don't forget to simplify your final answer as much as possible.

STEP 2

1. Identify the coefficients.
2. Apply the quadratic formula.
3. Simplify the result.

STEP 3

Let's define our quadratic equation as $ak^2 + bk + c = 0$ .
Looking at our given equation, $k^2 - k - 4 = 0$ , we can match it up to see what $a$ , $b$ , and $c$ are.

STEP 4

The coefficient of $k^2$ is $\bf{1}$ , so $a = \bf{1}$ .
The coefficient of $k$ is $\bf{-1}$ , so $b = \bf{-1}$ .
The constant term is $\bf{-4}$ , so $c = \bf{-4}$ .

STEP 5

Remember, the quadratic formula is given by: $k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Let's plug in our values for $a$ , $b$ , and $c$ .

STEP 6

Substituting $a = \bf{1}$ , $b = \bf{-1}$ , and $c = \bf{-4}$ , we get: $k = \frac{-(\bf{-1}) \pm \sqrt{(\bf{-1})^2 - 4 \cdot \bf{1} \cdot (\bf{-4})}}{2 \cdot \bf{1}}$

STEP 7

Let's simplify inside the square root first: $k = \frac{1 \pm \sqrt{1 + 16}}{2}$ $k = \frac{1 \pm \sqrt{17}}{2}$

STEP 8

Since $\sqrt{17}$ cannot be simplified further, our two solutions are: $k = \frac{1 + \sqrt{17}}{2} \quad \text{or} \quad k = \frac{1 - \sqrt{17}}{2}$

STEP 9

As decimals rounded to the nearest hundredth, these are approximately: $k \approx \frac{1 + 4.12}{2} \approx \frac{5.12}{2} \approx 2.56$ $k \approx \frac{1 - 4.12}{2} \approx \frac{-3.12}{2} \approx -1.56$

STEP 10

$k = \frac{1 + \sqrt{17}}{2}$ or $k = \frac{1 - \sqrt{17}}{2}$ . As decimals, $k \approx 2.56$ or $k \approx -1.56$ .

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QuestionSolve using the quadratic formula. $k^{2}-k-4=0$
Write your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth. $k=$ $\square$ or $k=$ $\square$ Submit

Studdy Solution

STEP 1

STEP 2

1. Identify the coefficients.
2. Apply the quadratic formula.
3. Simplify the result.

STEP 3

Let's define our quadratic equation as $ak^2 + bk + c = 0$ .
Looking at our given equation, $k^2 - k - 4 = 0$ , we can match it up to see what $a$ , $b$ , and $c$ are.

STEP 4

The coefficient of $k^2$ is $\bf{1}$ , so $a = \bf{1}$ .
The coefficient of $k$ is $\bf{-1}$ , so $b = \bf{-1}$ .
The constant term is $\bf{-4}$ , so $c = \bf{-4}$ .

STEP 5

Remember, the quadratic formula is given by: $k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Let's plug in our values for $a$ , $b$ , and $c$ .

STEP 6

Substituting $a = \bf{1}$ , $b = \bf{-1}$ , and $c = \bf{-4}$ , we get: $k = \frac{-(\bf{-1}) \pm \sqrt{(\bf{-1})^2 - 4 \cdot \bf{1} \cdot (\bf{-4})}}{2 \cdot \bf{1}}$

STEP 7

Let's simplify inside the square root first: $k = \frac{1 \pm \sqrt{1 + 16}}{2}$ $k = \frac{1 \pm \sqrt{17}}{2}$

STEP 8

Since $\sqrt{17}$ cannot be simplified further, our two solutions are: $k = \frac{1 + \sqrt{17}}{2} \quad \text{or} \quad k = \frac{1 - \sqrt{17}}{2}$

STEP 9

As decimals rounded to the nearest hundredth, these are approximately: $k \approx \frac{1 + 4.12}{2} \approx \frac{5.12}{2} \approx 2.56$ $k \approx \frac{1 - 4.12}{2} \approx \frac{-3.12}{2} \approx -1.56$

STEP 10

$k = \frac{1 + \sqrt{17}}{2}$ or $k = \frac{1 - \sqrt{17}}{2}$ . As decimals, $k \approx 2.56$ or $k \approx -1.56$ .

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QuestionSolve using the quadratic formula. k2−k−4=0k^{2}-k-4=0k2−k−4=0Write your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth. k=k=k= □\square□ or k=k=k= □\square□ Submit

Studdy Solution

STEP 1

STEP 2

1. Identify the coefficients.2. Apply the quadratic formula.3. Simplify the result.

STEP 3

Let's **define** our quadratic equation as ak2+bk+c=0ak^2 + bk + c = 0ak2+bk+c=0.Looking at our given equation, k2−k−4=0k^2 - k - 4 = 0k2−k−4=0, we can **match it up** to see what aaa, bbb, and ccc are.

STEP 4

The coefficient of k2k^2k2 is 1\bf{1}1, so a=1a = \bf{1}a=1.The coefficient of kkk is −1\bf{-1}−1, so b=−1b = \bf{-1}b=−1.The constant term is −4\bf{-4}−4, so c=−4c = \bf{-4}c=−4.

STEP 5

Remember, the **quadratic formula** is given by: k=−b±b2−4ac2ak = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}k=2a−b±b2−4ac​​ Let's **plug in** our values for aaa, bbb, and ccc.

STEP 6

Substituting a=1a = \bf{1}a=1, b=−1b = \bf{-1}b=−1, and c=−4c = \bf{-4}c=−4, we get: k=−(−1)±(−1)2−4⋅1⋅(−4)2⋅1k = \frac{-(\bf{-1}) \pm \sqrt{(\bf{-1})^2 - 4 \cdot \bf{1} \cdot (\bf{-4})}}{2 \cdot \bf{1}}k=2⋅1−(−1)±(−1)2−4⋅1⋅(−4)​​

STEP 7

Let's **simplify** inside the square root first: k=1±1+162k = \frac{1 \pm \sqrt{1 + 16}}{2}k=21±1+16​​ k=1±172k = \frac{1 \pm \sqrt{17}}{2}k=21±17​​

STEP 8

Since 17\sqrt{17}17​ cannot be simplified further, our **two solutions** are: k=1+172ork=1−172k = \frac{1 + \sqrt{17}}{2} \quad \text{or} \quad k = \frac{1 - \sqrt{17}}{2}k=21+17​​ork=21−17​​

STEP 9

STEP 10

k=1+172k = \frac{1 + \sqrt{17}}{2}k=21+17​​ or k=1−172k = \frac{1 - \sqrt{17}}{2}k=21−17​​. As decimals, k≈2.56k \approx 2.56k≈2.56 or k≈−1.56k \approx -1.56k≈−1.56.

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QuestionSolve using the quadratic formula. $k^{2}-k-4=0$
Write your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth. $k=$ $\square$ or $k=$ $\square$ Submit

1. Identify the coefficients.
2. Apply the quadratic formula.
3. Simplify the result.

Let's define our quadratic equation as $ak^2 + bk + c = 0$ .
Looking at our given equation, $k^2 - k - 4 = 0$ , we can match it up to see what $a$ , $b$ , and $c$ are.

The coefficient of $k^2$ is $\bf{1}$ , so $a = \bf{1}$ .
The coefficient of $k$ is $\bf{-1}$ , so $b = \bf{-1}$ .
The constant term is $\bf{-4}$ , so $c = \bf{-4}$ .

Remember, the quadratic formula is given by: $k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Let's plug in our values for $a$ , $b$ , and $c$ .

Substituting $a = \bf{1}$ , $b = \bf{-1}$ , and $c = \bf{-4}$ , we get: $k = \frac{-(\bf{-1}) \pm \sqrt{(\bf{-1})^2 - 4 \cdot \bf{1} \cdot (\bf{-4})}}{2 \cdot \bf{1}}$

Let's simplify inside the square root first: $k = \frac{1 \pm \sqrt{1 + 16}}{2}$ $k = \frac{1 \pm \sqrt{17}}{2}$

Since $\sqrt{17}$ cannot be simplified further, our two solutions are: $k = \frac{1 + \sqrt{17}}{2} \quad \text{or} \quad k = \frac{1 - \sqrt{17}}{2}$

$k = \frac{1 + \sqrt{17}}{2}$ or $k = \frac{1 - \sqrt{17}}{2}$ . As decimals, $k \approx 2.56$ or $k \approx -1.56$ .