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Solve using the quadratic formula. $$3 x^{2}+5 x=-11$$

Short Answer

Expert verified
The solutions are complex: \(x = \frac{-5 \pm i\sqrt{107}}{6}\).

Step by step solution

01

Move all terms to one side of the equation

To use the quadratic formula, the equation must be in the standard form: \( ax^2 + bx + c = 0 \). Let's move all terms to one side to get: \(3x^2 + 5x + 11 = 0\).
02

Identify coefficients

Identify the coefficients \(a\), \(b\), and \(c\) from the quadratic equation \(3x^2 + 5x + 11 = 0\). Here, \(a = 3\), \(b = 5\), and \(c = 11\).
03

Write down the quadratic formula

The quadratic formula is \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\). We will use this to solve for \(x\).
04

Substitute the values into the formula

Substitute \(a = 3\), \(b = 5\), and \(c = 11\) into the quadratic formula: \[x = \frac{{-(5) \pm \sqrt{{(5)^2 - 4(3)(11)}}}}{2(3)}\]
05

Simplify inside the square root

Calculate the discriminant, \(b^2 - 4ac\):\[(5)^2 - 4(3)(11) = 25 - 132 = -107\]Since the discriminant is negative, the solutions will involve complex numbers.
06

Calculate the solutions

Substitute the discriminant back into the quadratic formula:\[x = \frac{{-5 \pm \sqrt{{-107}}}}{6}\]This can be rewritten using imaginary numbers:\[x = \frac{-5 \pm i\sqrt{107}}{6}\]
07

Express the solution

The solutions are\[x = \frac{-5 + i\sqrt{107}}{6}\quad \text{and}\quad x = \frac{-5 - i\sqrt{107}}{6}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complex Numbers
Complex numbers are a fascinating aspect of algebra that allow us to find solutions to equations that cannot be solved solely by real numbers. A complex number is expressed in the form \(a + bi\), where \(a\) is the real part and \(bi\) is the imaginary part. These numbers are essential for equations where the square root of a negative number is encountered.
In the quadratic equation provided, the discriminant is negative, indicating no real solutions exist. This is where complex numbers come in handy.
  • The quadratic formula gives us complex solutions when the discriminant is less than zero.
  • Using \(i = \sqrt{-1}\), which is the imaginary unit, we can express solutions in terms of \(i\).
By converting negative square roots into terms involving \(i\), mathematicians developed a more comprehensive number system. This system extends the real numbers to include complex numbers, allowing deeper exploration of algebraic solutions.
Discriminant
The discriminant is a key concept in understanding the nature of the roots of a quadratic equation. It's found within the quadratic formula as \(b^2 - 4ac\). By examining the sign of the discriminant, we can determine the types of roots an equation will have.
  • If the discriminant is positive, there are two distinct real roots.
  • If it's zero, there's exactly one real root, sometimes referred to as a repeated root.
  • A negative discriminant, as in our problem, indicates two complex roots.
The discriminant acts as a diagnostic tool in solving quadratic equations. In the example given, the discriminant \(b^2 - 4ac\) equals \(-107\). This directly shows that the solutions must involve complex numbers, broadening our understanding of the equation's solvability.
Imaginary Numbers
Imaginary numbers might sound perplexing, but they were born out of a necessity to solve problems that real numbers couldn't handle. The imaginary unit \(i\) is defined as \(i = \sqrt{-1}\).
  • This allows us to work with square roots of negative numbers by transforming them into a product of \(i\) and a positive number.
  • In the quadratic solution involving \(\sqrt{-107}\), we convert it to \(i\sqrt{107}\), thus expressing a value computable in terms of real and imaginary components.
Using the squared property of \(i\) where \(i^2 = -1\), we can manipulate and work with these numbers in algebraic calculations. Imaginary numbers, combined with real numbers, enable us to represent complex numbers, providing a complete set of solutions to quadratic equations without any limitations due to negative square roots.

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