Solve the Quadratic Equation: x² + 5x + 4 = 0

Quadratic Equations with Factoring Method

Solve the following problem:

x2+5x+4=0 x^2+5x+4=0

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Solve
00:04 Let's use the root formula
00:31 Let's identify the coefficients
00:41 We'll substitute appropriate values according to the given data to find the solutions
01:17 Let's calculate the square and products
01:32 Let's calculate the square root of 9
01:43 We'll find the 2 possible solutions (addition, subtraction)
01:58 This is one solution, let's continue to solve the second one
02:11 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 problem:

x2+5x+4=0 x^2+5x+4=0

2

Step-by-step solution

This is a quadratic equation:

x2+5x+4=0=0 x^2+5x+4=0 =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 in to a form where all 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+4=0=0 x^2+5x+4=0 =0

And solve it:

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

{a=1b=5c=4 \begin{cases}a=1 \\ b=5 \\ c=4\end{cases}

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

We obtain the solutions of the equation (its roots) by insertion we just identified into the quadratic formula:

x1,2=b±b24ac2a=5±5241421 x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-5\pm\sqrt{5^2-4\cdot1\cdot4}}{2\cdot1}

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

x1,2=5±92=5±32 x_{1,2}=\frac{-5\pm\sqrt{9}}{2}=\frac{-5\pm3}{2}

Therefore the solutions of the equation are:

{x1=5+32=1x2=532=4 \begin{cases}x_1=\frac{-5+3}{2}=-1 \\ x_2=\frac{-5-3}{2}=-4\end{cases}

Therefore the correct answer is answer C.

3

Final Answer

x1=1,x2=4 x_1=-1,x_2=-4

Key Points to Remember

Essential concepts to master this topic
  • Standard Form: Arrange equation as ax² + bx + c = 0
  • Quadratic Formula: x = (-5 ± √(25-16))/2 = (-5 ± 3)/2
  • Verification: Check both solutions: (-1)² + 5(-1) + 4 = 0 ✓

Common Mistakes

Avoid these frequent errors
  • Forgetting the ± symbol in quadratic formula
    Don't write just x = (-b + √discriminant)/2a = only one solution! This gives you only half the answer since quadratic equations have two solutions. Always use the ± symbol to find both x₁ = (-b + √discriminant)/2a and x₂ = (-b - √discriminant)/2a.

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 does a quadratic equation have two solutions?

+

A quadratic equation represents a parabola that crosses the x-axis at two points (usually). Each crossing point gives you one solution, so you get two x-values where the equation equals zero.

When should I use the quadratic formula vs factoring?

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Try factoring first if you can easily find two numbers that multiply to give 'c' and add to give 'b'. If factoring seems difficult, the quadratic formula always works for any quadratic equation!

What if I get a negative number under the square root?

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If the discriminant (b² - 4ac) is negative, the equation has no real solutions. This means the parabola doesn't cross the x-axis at any real points.

How do I remember the quadratic formula?

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Try this song: "x equals negative b, plus or minus the square root, of b squared minus 4ac, all over 2a!" Practice writing it out several times too.

Can I check my work without substituting back?

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You can verify using Vieta's formulas: the sum of roots should equal -b/a, and the product should equal c/a. For x2+5x+4=0 x^2+5x+4=0 : (-1) + (-4) = -5 ✓ and (-1)×(-4) = 4 ✓

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