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

Question

Solve the following equation:

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

Video Solution

Solution Steps

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 Solution

Notice that the quadratic equation:

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

and this is because there is a quadratic term (meaning raised to the second power),

The first step in solving a quadratic equation is always arranging it in a form where all terms on one side are ordered from highest to 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 with the solving technique:

We'll choose to solve it using the quadratic formula,

Let's recall it first:

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 get a true statement when substituted in 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,

And we'll get the equation's solutions (roots) by substituting 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 and calculate the expression inside the square root and simplify the expression:

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

Therefore 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.

Answer

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