Solving Quadratic Equations Using The Quadratic Formula Worksheet - Free Printable
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Step-by-step solution for: Solving Quadratic Equations Using The Quadratic Formula Worksheet
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Show Answer Key & Explanations
Step-by-step solution for: Solving Quadratic Equations Using The Quadratic Formula Worksheet
To solve quadratic equations using the quadratic formula, we use the formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where the quadratic equation is in the standard form \( ax^2 + bx + c = 0 \).
Let's solve each equation step by step.
---
Here, \( a = 3 \), \( b = -5 \), and \( c = -8 \).
\[
n = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(-8)}}{2(3)}
\]
\[
n = \frac{5 \pm \sqrt{25 + 96}}{6}
\]
\[
n = \frac{5 \pm \sqrt{121}}{6}
\]
\[
n = \frac{5 \pm 11}{6}
\]
This gives two solutions:
\[
n = \frac{5 + 11}{6} = \frac{16}{6} = \frac{8}{3}
\]
\[
n = \frac{5 - 11}{6} = \frac{-6}{6} = -1
\]
So, the solutions are:
\[
n = \frac{8}{3}, \quad n = -1
\]
---
Here, \( a = 1 \), \( b = 10 \), and \( c = 21 \).
\[
x = \frac{-10 \pm \sqrt{10^2 - 4(1)(21)}}{2(1)}
\]
\[
x = \frac{-10 \pm \sqrt{100 - 84}}{2}
\]
\[
x = \frac{-10 \pm \sqrt{16}}{2}
\]
\[
x = \frac{-10 \pm 4}{2}
\]
This gives two solutions:
\[
x = \frac{-10 + 4}{2} = \frac{-6}{2} = -3
\]
\[
x = \frac{-10 - 4}{2} = \frac{-14}{2} = -7
\]
So, the solutions are:
\[
x = -3, \quad x = -7
\]
---
Here, \( a = 10 \), \( b = -9 \), and \( c = 6 \).
\[
x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(10)(6)}}{2(10)}
\]
\[
x = \frac{9 \pm \sqrt{81 - 240}}{20}
\]
\[
x = \frac{9 \pm \sqrt{-159}}{20}
\]
Since the discriminant (\( b^2 - 4ac \)) is negative, there are no real solutions. The solutions are complex:
\[
x = \frac{9 \pm i\sqrt{159}}{20}
\]
So, the solutions are:
\[
x = \frac{9 + i\sqrt{159}}{20}, \quad x = \frac{9 - i\sqrt{159}}{20}
\]
---
This can be rewritten as:
\[
p^2 = 9
\]
Taking the square root of both sides:
\[
p = \pm 3
\]
So, the solutions are:
\[
p = 3, \quad p = -3
\]
---
Here, \( a = 6 \), \( b = -12 \), and \( c = 1 \).
\[
x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(6)(1)}}{2(6)}
\]
\[
x = \frac{12 \pm \sqrt{144 - 24}}{12}
\]
\[
x = \frac{12 \pm \sqrt{120}}{12}
\]
\[
x = \frac{12 \pm 2\sqrt{30}}{12}
\]
\[
x = \frac{6 \pm \sqrt{30}}{6}
\]
So, the solutions are:
\[
x = \frac{6 + \sqrt{30}}{6}, \quad x = \frac{6 - \sqrt{30}}{6}
\]
---
This can be rewritten as:
\[
6n^2 = 11
\]
\[
n^2 = \frac{11}{6}
\]
Taking the square root of both sides:
\[
n = \pm \sqrt{\frac{11}{6}}
\]
So, the solutions are:
\[
n = \sqrt{\frac{11}{6}}, \quad n = -\sqrt{\frac{11}{6}}
\]
---
Here, \( a = 2 \), \( b = 5 \), and \( c = -9 \).
\[
n = \frac{-5 \pm \sqrt{5^2 - 4(2)(-9)}}{2(2)}
\]
\[
n = \frac{-5 \pm \sqrt{25 + 72}}{4}
\]
\[
n = \frac{-5 \pm \sqrt{97}}{4}
\]
So, the solutions are:
\[
n = \frac{-5 + \sqrt{97}}{4}, \quad n = \frac{-5 - \sqrt{97}}{4}
\]
---
Here, \( a = 3 \), \( b = -6 \), and \( c = -23 \).
\[
x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(-23)}}{2(3)}
\]
\[
x = \frac{6 \pm \sqrt{36 + 276}}{6}
\]
\[
x = \frac{6 \pm \sqrt{312}}{6}
\]
\[
x = \frac{6 \pm 2\sqrt{78}}{6}
\]
\[
x = \frac{3 \pm \sqrt{78}}{3}
\]
So, the solutions are:
\[
x = \frac{3 + \sqrt{78}}{3}, \quad x = \frac{3 - \sqrt{78}}{3}
\]
---
\[
\boxed{
\begin{aligned}
1. & \quad n = \frac{8}{3}, \, n = -1 \\
2. & \quad x = -3, \, x = -7 \\
3. & \quad x = \frac{9 + i\sqrt{159}}{20}, \, x = \frac{9 - i\sqrt{159}}{20} \\
4. & \quad p = 3, \, p = -3 \\
5. & \quad x = \frac{6 + \sqrt{30}}{6}, \, x = \frac{6 - \sqrt{30}}{6} \\
6. & \quad n = \sqrt{\frac{11}{6}}, \, n = -\sqrt{\frac{11}{6}} \\
7. & \quad n = \frac{-5 + \sqrt{97}}{4}, \, n = \frac{-5 - \sqrt{97}}{4} \\
8. & \quad x = \frac{3 + \sqrt{78}}{3}, \, x = \frac{3 - \sqrt{78}}{3}
\end{aligned}
}
\]
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where the quadratic equation is in the standard form \( ax^2 + bx + c = 0 \).
Let's solve each equation step by step.
---
1. \( 3n^2 - 5n - 8 = 0 \)
Here, \( a = 3 \), \( b = -5 \), and \( c = -8 \).
\[
n = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(-8)}}{2(3)}
\]
\[
n = \frac{5 \pm \sqrt{25 + 96}}{6}
\]
\[
n = \frac{5 \pm \sqrt{121}}{6}
\]
\[
n = \frac{5 \pm 11}{6}
\]
This gives two solutions:
\[
n = \frac{5 + 11}{6} = \frac{16}{6} = \frac{8}{3}
\]
\[
n = \frac{5 - 11}{6} = \frac{-6}{6} = -1
\]
So, the solutions are:
\[
n = \frac{8}{3}, \quad n = -1
\]
---
2. \( x^2 + 10x + 21 = 0 \)
Here, \( a = 1 \), \( b = 10 \), and \( c = 21 \).
\[
x = \frac{-10 \pm \sqrt{10^2 - 4(1)(21)}}{2(1)}
\]
\[
x = \frac{-10 \pm \sqrt{100 - 84}}{2}
\]
\[
x = \frac{-10 \pm \sqrt{16}}{2}
\]
\[
x = \frac{-10 \pm 4}{2}
\]
This gives two solutions:
\[
x = \frac{-10 + 4}{2} = \frac{-6}{2} = -3
\]
\[
x = \frac{-10 - 4}{2} = \frac{-14}{2} = -7
\]
So, the solutions are:
\[
x = -3, \quad x = -7
\]
---
3. \( 10x^2 - 9x + 6 = 0 \)
Here, \( a = 10 \), \( b = -9 \), and \( c = 6 \).
\[
x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(10)(6)}}{2(10)}
\]
\[
x = \frac{9 \pm \sqrt{81 - 240}}{20}
\]
\[
x = \frac{9 \pm \sqrt{-159}}{20}
\]
Since the discriminant (\( b^2 - 4ac \)) is negative, there are no real solutions. The solutions are complex:
\[
x = \frac{9 \pm i\sqrt{159}}{20}
\]
So, the solutions are:
\[
x = \frac{9 + i\sqrt{159}}{20}, \quad x = \frac{9 - i\sqrt{159}}{20}
\]
---
4. \( p^2 - 9 = 0 \)
This can be rewritten as:
\[
p^2 = 9
\]
Taking the square root of both sides:
\[
p = \pm 3
\]
So, the solutions are:
\[
p = 3, \quad p = -3
\]
---
5. \( 6x^2 - 12x + 1 = 0 \)
Here, \( a = 6 \), \( b = -12 \), and \( c = 1 \).
\[
x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(6)(1)}}{2(6)}
\]
\[
x = \frac{12 \pm \sqrt{144 - 24}}{12}
\]
\[
x = \frac{12 \pm \sqrt{120}}{12}
\]
\[
x = \frac{12 \pm 2\sqrt{30}}{12}
\]
\[
x = \frac{6 \pm \sqrt{30}}{6}
\]
So, the solutions are:
\[
x = \frac{6 + \sqrt{30}}{6}, \quad x = \frac{6 - \sqrt{30}}{6}
\]
---
6. \( 6n^2 - 11 = 0 \)
This can be rewritten as:
\[
6n^2 = 11
\]
\[
n^2 = \frac{11}{6}
\]
Taking the square root of both sides:
\[
n = \pm \sqrt{\frac{11}{6}}
\]
So, the solutions are:
\[
n = \sqrt{\frac{11}{6}}, \quad n = -\sqrt{\frac{11}{6}}
\]
---
7. \( 2n^2 + 5n - 9 = 0 \)
Here, \( a = 2 \), \( b = 5 \), and \( c = -9 \).
\[
n = \frac{-5 \pm \sqrt{5^2 - 4(2)(-9)}}{2(2)}
\]
\[
n = \frac{-5 \pm \sqrt{25 + 72}}{4}
\]
\[
n = \frac{-5 \pm \sqrt{97}}{4}
\]
So, the solutions are:
\[
n = \frac{-5 + \sqrt{97}}{4}, \quad n = \frac{-5 - \sqrt{97}}{4}
\]
---
8. \( 3x^2 - 6x - 23 = 0 \)
Here, \( a = 3 \), \( b = -6 \), and \( c = -23 \).
\[
x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(-23)}}{2(3)}
\]
\[
x = \frac{6 \pm \sqrt{36 + 276}}{6}
\]
\[
x = \frac{6 \pm \sqrt{312}}{6}
\]
\[
x = \frac{6 \pm 2\sqrt{78}}{6}
\]
\[
x = \frac{3 \pm \sqrt{78}}{3}
\]
So, the solutions are:
\[
x = \frac{3 + \sqrt{78}}{3}, \quad x = \frac{3 - \sqrt{78}}{3}
\]
---
Final Answer
\[
\boxed{
\begin{aligned}
1. & \quad n = \frac{8}{3}, \, n = -1 \\
2. & \quad x = -3, \, x = -7 \\
3. & \quad x = \frac{9 + i\sqrt{159}}{20}, \, x = \frac{9 - i\sqrt{159}}{20} \\
4. & \quad p = 3, \, p = -3 \\
5. & \quad x = \frac{6 + \sqrt{30}}{6}, \, x = \frac{6 - \sqrt{30}}{6} \\
6. & \quad n = \sqrt{\frac{11}{6}}, \, n = -\sqrt{\frac{11}{6}} \\
7. & \quad n = \frac{-5 + \sqrt{97}}{4}, \, n = \frac{-5 - \sqrt{97}}{4} \\
8. & \quad x = \frac{3 + \sqrt{78}}{3}, \, x = \frac{3 - \sqrt{78}}{3}
\end{aligned}
}
\]
Parent Tip: Review the logic above to help your child master the concept of solving quadratic equations worksheet.
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