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Use the quadratic formula to solve. Express your answer in simplest form.
25, k, squared, plus, 36, k, plus, 9, equals, 6, k
25k
2
+36k+9=
6k
Answer
Attempt 1 out of 5
Additional Solution
No Solution
k, equalsk=
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square root of, end square root
x
plus or minus
±
Watch Video
Show Examples
Use the quadratic formula to solve. Express your answer in simplest form.
25, k, squared, plus, 36, k, plus, 9, equals, 6, k
25k
2
+36k+9=
6k
Answer
Attempt 1 out of 5
Additional Solution
No Solution
k, equalsk=
Submit Answer
square root of, end square root
x
plus or minus
±
Answers
Answered by
GPT-4o mini
To solve the equation \( 25k^2 + 36k + 9 = 6k \) using the quadratic formula, we first need to rearrange it into standard form \( ax^2 + bx + c = 0 \).
Starting with the equation:
\[
25k^2 + 36k + 9 = 6k
\]
We can subtract \( 6k \) from both sides:
\[
25k^2 + 36k - 6k + 9 = 0
\]
Which simplifies to:
\[
25k^2 + 30k + 9 = 0
\]
Now, we can identify the coefficients:
- \( a = 25 \)
- \( b = 30 \)
- \( c = 9 \)
The quadratic formula is given by:
\[
k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Now, we will calculate the discriminant \( b^2 - 4ac \):
\[
b^2 = (30)^2 = 900
\]
\[
4ac = 4 \times 25 \times 9 = 900
\]
\[
b^2 - 4ac = 900 - 900 = 0
\]
Since the discriminant is zero, there is one real solution:
\[
k = \frac{-b \pm \sqrt{0}}{2a} = \frac{-30 \pm 0}{2 \times 25}
\]
\[
k = \frac{-30}{50} = -\frac{3}{5}
\]
Thus, the solution is:
\[
\boxed{-\frac{3}{5}}
\]
Starting with the equation:
\[
25k^2 + 36k + 9 = 6k
\]
We can subtract \( 6k \) from both sides:
\[
25k^2 + 36k - 6k + 9 = 0
\]
Which simplifies to:
\[
25k^2 + 30k + 9 = 0
\]
Now, we can identify the coefficients:
- \( a = 25 \)
- \( b = 30 \)
- \( c = 9 \)
The quadratic formula is given by:
\[
k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Now, we will calculate the discriminant \( b^2 - 4ac \):
\[
b^2 = (30)^2 = 900
\]
\[
4ac = 4 \times 25 \times 9 = 900
\]
\[
b^2 - 4ac = 900 - 900 = 0
\]
Since the discriminant is zero, there is one real solution:
\[
k = \frac{-b \pm \sqrt{0}}{2a} = \frac{-30 \pm 0}{2 \times 25}
\]
\[
k = \frac{-30}{50} = -\frac{3}{5}
\]
Thus, the solution is:
\[
\boxed{-\frac{3}{5}}
\]
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