Solve x² + 5x + 6 = 0: Complete Quadratic Equation Guide

Quadratic Equations with Factoring Method

Solve the following equation:

x2+5x+6=0 x^2+5x+6=0

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Find X
00:03 Use the roots formula
00:29 Identify the coefficients
00:39 Substitute appropriate values according to the given data and solve
01:06 Calculate the square and products
01:28 One root is always equal to 1
01:35 These are the 2 possible solutions (addition,subtraction)
01:51 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Solve the following equation:

x2+5x+6=0 x^2+5x+6=0

2

Step-by-step solution

This is a quadratic equation:

x2+5x+6=0 x^2+5x+6=0

due to the fact that there is a quadratic term (meaning raised to the second power),

The first step in solving a quadratic equation is always arranging it to a form where all the terms on one side are ordered from the highest to the lowest power (in descending order from left to right) and 0 on the other side,

Then we can choose whether to solve it using the quadratic formula or by factoring/completing the square.

The equation in the problem is already arranged, so let's proceed to solve it using the quadratic formula,

Remember:

The rule states that the roots of an equation of the form:

ax2+bx+c=0 ax^2+bx+c=0

are:

x1,2=b±b24ac2a x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

(meaning its solutions, the two possible values of the unknown for which we obtain a true statement when inserted into the equation)

This formula is called: "The Quadratic Formula"

Let's return to the problem:

x2+5x+6=0 x^2+5x+6=0 and solve it:

First, let's identify the coefficients of the terms:

{a=1b=5c=6 \begin{cases}a=1 \\ b=5 \\ c=6\end{cases}

where we noted that the coefficient of the quadratic term is 1,

We obtain the equation's solutions (roots) by inserting the coefficients we just noted into the quadratic formula:

x1,2=b±b24ac2a=5±5241621 x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-5\pm\sqrt{5^2-4\cdot1\cdot6}}{2\cdot1}

Let's continue to calculate the expression inside of the square root and proceed to simplify the expression:

x1,2=5±12=5±12 x_{1,2}=\frac{-5\pm\sqrt{1}}{2}=\frac{-5\pm1}{2}

The solutions to the equation are:

{x1=5+12=2x2=512=3 \begin{cases}x_1=\frac{-5+1}{2}=-2 \\ x_2=\frac{-5-1}{2}=-3\end{cases}

Therefore the correct answer is answer D.

3

Final Answer

x1=3,x2=2 x_1=-3,x_2=-2

Key Points to Remember

Essential concepts to master this topic
  • Standard Form: Arrange as ax² + bx + c = 0 before solving
  • Quadratic Formula: x = (-b ± √(b² - 4ac))/2a where a=1, b=5, c=6
  • Check: Substitute x = -2: (-2)² + 5(-2) + 6 = 0 ✓

Common Mistakes

Avoid these frequent errors
  • Sign errors when applying quadratic formula
    Don't forget the negative sign in front of b = -(-5) = +5! This leads to wrong solutions like x = 3, -2 instead of x = -3, -2. Always write out x = (-b ± √(discriminant))/2a with careful attention to all signs.

Practice Quiz

Test your knowledge with interactive questions

a = Coefficient of x²

b = Coefficient of x

c = Coefficient of the independent number


what is the value of \( a \) in the equation

\( y=3x-10+5x^2 \)

FAQ

Everything you need to know about this question

Why do we get two solutions for quadratic equations?

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Quadratic equations create parabolas when graphed, and these curves can cross the x-axis at two points. Each crossing point represents a solution where the equation equals zero!

Can I solve this by factoring instead of using the formula?

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Yes! Since x2+5x+6=(x+2)(x+3) x^2 + 5x + 6 = (x + 2)(x + 3) , you get the same solutions: x = -2 and x = -3. Factoring is often faster when it works easily.

What does the discriminant b² - 4ac tell me?

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The discriminant b24ac=2524=1 b^2 - 4ac = 25 - 24 = 1 tells you about solutions:

  • Positive: Two real solutions
  • Zero: One repeated solution
  • Negative: No real solutions

How do I remember the quadratic formula?

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Try this rhyme: "x equals negative b, plus or minus the square root, of b squared minus 4ac, all over 2a!" Practice writing it several times to build muscle memory.

Should I always use the quadratic formula?

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Not always! Try factoring first if the numbers look friendly. Use the quadratic formula when factoring seems difficult or when you want to be absolutely sure of your answer.

What if I get decimal solutions?

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That's completely normal! Many quadratic equations have irrational or decimal solutions. Just make sure to check your arithmetic and verify by substituting back into the original equation.

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