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Solve by using the quadratic formula. $$x^{2}+6 x+3=0$$

Short Answer

Expert verified
The solutions are \( x = -3 + \sqrt{6} \) and \( x = -3 - \sqrt{6} \).

Step by step solution

01

Identify the coefficients

In order to solve the quadratic equation using the quadratic formula, we must first identify the coefficients from the equation: The standard form of a quadratic equation is: \[ ax^2 + bx + c = 0 \]In the given equation, \( x^2 + 6x + 3 = 0 \), - \( a = 1 \)- \( b = 6 \)- \( c = 3 \).
02

Write down the quadratic formula

The quadratic formula used to find the solutions of the equation \( ax^2 + bx + c = 0 \) is:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
03

Substitute the coefficients into the formula

Substitute the values of \( a \), \( b \), and \( c \) into the quadratic formula:\[ x = \frac{-6 \pm \sqrt{6^2 - 4 imes 1 imes 3}}{2 imes 1} \]
04

Calculate the discriminant

Calculate the value under the square root (the discriminant):\[ b^2 - 4ac = 6^2 - 4 imes 1 imes 3 = 36 - 12 = 24 \]
05

Solve the equation

Now, insert the discriminant back into the formula and solve for the two potential values of \( x \):\[ x = \frac{-6 \pm \sqrt{24}}{2} \]\[ \sqrt{24} = 2\sqrt{6} \] (since 24 can be simplified to \( 4 imes 6 \) and \( \sqrt{4} = 2 \))\[ x = \frac{-6 \pm 2\sqrt{6}}{2} \]
06

Simplify the expression

Divide each term in the numerator by 2 to simplify:\[ x = \frac{-6}{2} \pm \frac{2\sqrt{6}}{2} \]\[ x = -3 \pm \sqrt{6} \]This gives the two solutions:\[ x = -3 + \sqrt{6} \] and \[ x = -3 - \sqrt{6} \]

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Key Concepts

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

quadratic equation
A quadratic equation is a type of polynomial equation that involves a variable raised to the second power. The general form of a quadratic equation is \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). This means that the highest exponent of the variable \( x \) is 2. Quadratic equations are fundamental in algebra and appear often in various real-world applications.
  • The term \( ax^2 \) is the quadratic term.
  • The term \( bx \) is the linear term.
  • The term \( c \) is the constant term.
Understanding the form and structure of quadratic equations is crucial for solving them using different methods.
discriminant
The discriminant is a key component of the quadratic formula and helps us determine the nature of the solutions to a quadratic equation. It is found under the square root in the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). The discriminant is the expression \( b^2 - 4ac \).
The value of the discriminant tells us:
  • If \( b^2 - 4ac > 0 \), there are two distinct real solutions.
  • If \( b^2 - 4ac = 0 \), there is exactly one real solution (the solutions are repeated).
  • If \( b^2 - 4ac < 0 \), there are two complex (not real) solutions.
In our equation \( x^2 + 6x + 3 = 0 \), we calculated the discriminant as 24. Since 24 is greater than zero, we have two distinct real solutions.
solutions of a quadratic equation
The solutions of a quadratic equation are the values of \( x \) that satisfy the equation \( ax^2 + bx + c = 0 \). These solutions can be found using several methods, including factoring, completing the square, or using the quadratic formula.
Using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), provides a direct way of finding solutions, especially when the equation doesn't factor easily. In our example, substituting \( a = 1 \), \( b = 6 \), \( c = 3 \) into the quadratic formula gives us:\[ x = \frac{-6 \pm \sqrt{24}}{2} \]
After simplifying, the solutions are:
  • \( x = -3 + \sqrt{6} \)
  • \( x = -3 - \sqrt{6} \)
These solutions indicate the points where the quadratic equation would cross or touch the \( x \)-axis if graphed.
simplifying radicals
Simplifying radicals is an important step in solving quadratic equations when dealing with the square root part of the quadratic formula. It involves breaking down a radical expression into its simplest form.
For instance, with the discriminant \( \sqrt{24} \), we simplify as follows:
  • The number 24 can be factored into \( 4 \times 6 \).
  • Since \( \sqrt{4} = 2 \), the expression becomes \( 2\sqrt{6} \).
By simplifying radicals, we make calculations cleaner and easier to interpret. In our equation, simplifying \( \sqrt{24} \) into \( 2\sqrt{6} \) allowed us to further simplify each component of the quadratic formula, resulting in the final solutions: \( x = -3 \pm \sqrt{6} \).

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